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The  Algebraic  Solution  of  Equations  ^' 

OF    ANY.  DEGREE 


A   Novel,  Simple  and]  Direct  Method  for  the 
Solution  of  Equations  of  the  N^^  Degree 


BY 


L.  A.  BUCHANAN,  M.  E. 

Instructor  in  Industrial  Education  for  the  City  of  Stockton.    Former 

Principal  of  the  Cogswell  Polytechnic  College,  Instructor  in  the 

Shop  Work  in  the  Leland  Stanford  Junior  University. 


AND      h    yjsiivEKSlTY 

J.  LEWIS    ANDFf^^^Ss^ 

Graduate  of  the  Polytechnic  High  School  and  the  Cogswell  Polytechnic 
College,  San  Francisco,  Cal. 


PUBLISHED  FOR  THE  AUTHORS 
BY 

THE  WHITAKER   &  RAY  COMPANY 
SAN   FRANCISCO,  CAL. 


Copyrighted  1899 

BY 
A.  A.  ANDRK 


PREFACE. 


^j 


In  presenting  the  following  pages  we  are  attempting  to 
introduce  a  method  for  the  solution  of  equations  of  any 
degree  which  we  believe  to  be  novel  and  simple.  A  method 
that  resorts  to  no  artificial  nor  indirect  processes  for  the  com- 
pletion of  the  given  quantity  to  produce  perfect  squares ;  nor 
any  mathematical  substitution  or  jugglery^that  may  lead  the 
student  to  fields  not  strictly  algebraic,  involving  trigonometry, 
graphics,  or  the  various  processes  sometimes  resorted  to  in 

the  solution  of  equations  of  a  higher  degree  than  the  quad-  ^   | 

ratic.^  *>^   J 

We  believe  the  method  to  be  direct,  simple  and  strictly  in  \^  l| 

accordance  with  the  common  sense  dictates   of   the    given  ^    v 

equation,  to  be  solved  ;  to  illustrate,  if  bxw=^c  and  the  value  >k    i 

of  X  be  required  ;  we  divide  c  by   the  coefficient  of  x  and  -^    -I 

extract  the  n^  root  of  the  unknown  quantity   and    so    on  ^    '^ 

throughout  for  any  given  power  or  number  of  terms  when  1^     ^ 

the  exponents  are  integers  or  can   be  readily  transformed  to  ''^^   v. 

integral  powers.  **    -J 

The  first  chapter  is  devoted  to  a  comprehensive  review  on  5 

the  binomial  formula  upon  which  the  solution  of  the  method  ^^ 

herein  given  is  solely  based,  and  the   processes  of  involution  *"  | 

and  evolution.     These    topics    have  been    dealt  with   thor-  i 

oughly  so  far  as  it  has  been  thought  necessary  and  helpful  to  .^ 

explain  this  method.  >t 

In  the  chapter  on  equations,  we  have  added  some  of  the  ^ 

the  theory  of  equations,   such  as  Descartes  Rule,   and    the  ^ 

theory  showing  the  relation  between  the   coefficient  of/  (jt:)  j 

and  the  roots  of  the  equation  /  (^x)  =  0:  and  as  much  more  ^ 

as  was  thought  useful  and  necessary.  ^ 

100682 


iv  Preface, 

We  have  given  the  complete  solution  of  quite  a  number  of 
examples  that  differ  from  each  other  only  by  slight  modifica- 
tions of  sign  or  degree  ;  but  experience  has  taught  us  that 
the  average  student  learns  more  by  following  throughout  an 
actual  construction,  than  by  the  language  of  demonstration 
and  rule ;  hence  we  have  attempted  to  show  all  modifications 
of  sign,  degree  and  quantity  that  may  come  under  general 
observation  and  practice. 

It  is  hoped  that  the  following  method  will  be  found  useful 
to  the  engineer  and  man  of  practice  as  well  as  the  student  ;  for 
the  ordinary  methods  of  solving,  even  an  equation  of  the 
third  degree,  are  laborous  and  approximate,  involving  graphic 
constructions  or  the  denser  analytical  methods  of  Sturm  and 
Horner. 

While  rules  for  the  application  of  the  method  are  given  in 
these  pages,  yet  we  have  also  attempted  to  make  these  rules, 
laws,  or  a  deduction  from  a  process  of  reasoning,  and  not 
merely  statements  to  follow  blindly. 

We  have  consulted  the  works  of  many  authors  on  algebra 
and  used  such  demonstration  from  each  as  we  considered 
useful. 

Respectfully, 

THE  AUTHORS. 


The  Solution  ofTlgebraic  Eouations 


OF    ANY    DEGKEE. 


CHAPTER  I. 
BINOMIAL    FORMULA. 


For  the  sake  of  a  clear  understanding  of  what  follows,  we 
shall  develop  the  Binomial  Formula  upon  which  the  theory 
of  the  following  demonstration   depends. 


(x+a)  (x+5)=x''  +  a 


x+ad 


(x+a)  (x+d)  (x+c)=x^-\-a 


x^'+ad 
+  ac 
+  dc 


x+adc 


{x+a)(x+d){x+c)(x+d)=:x^  +  a 


+  c\ 
+  d\ 


x^+adc 
+  abd 
+  acd 
+  dcd 


X  +  adcd 


'x^  +  ad. 

+  ac 

+  ad 

+  dc 
'     +dd 

From  the  above  we  observe  the  following  laws  : 

First — The  nujnber  of  terms  in  the  second  member  is 

one  greater   than  the  number  of  binomial  factors  in  the 

first  member. 

Second  — 7"/^^  exponent  of  a  in  the  first  term  of  the  second 
member  is  equal  to  the  number  of  binomial  factors,  and  in 
each  of  the  succeeding  terms  the  exponent  of  x  is  one  less 
than  in  the  preceding  term. 


2  Binomial  Formula. 

Third —  The  coefficient  of  the  first  term  of  the  second 
member  is  unity,  the  coefficient  of  the  second  term  is  the 
sum  of  the  second  terms  of  the  binomial  factors ;  the 
coefficient  of  the  third  term  is  the  sum  of  all  the  products 
of  the  second  terms  of  the  binomial  factors^  taken  two  at  a 
time  ;  the  coefficient  of  the  fourth  term  is  the  sum  of  all  the 
products  of  the  second  terms  of  the  binomial  factors,  taken 
three  at  a  time,  and  so  on;  the  last  term  is  the  product  of 
all  the  second  terms  of  the  binomial  factors. 

Let  us  assume  it  to  be  true  for  n  binomials, 

{x-^-a)  {x-\-b)  {x-k-c)  {x-\-d^ {x-\-k)  therefore  ; 

{x^-d)   (^+^)  {x-^c~)  {x\d)  ....   {X'\-k)=x'^-\'Px''-^ 

+P^x''-^+P^x''-^+ +P»,  where    P=  the   sum 

of  the  terms  a,  b,  c,  d,,  , .  ,k.     (1) 

/>^=the  sum  of  the  products  of  these  terms,  taken  two  at  a  time. 

/>«=the  product  of  all  these  terms. 

Multiply   both  members  of  (1)   by  (x+l)  and  from  the 

above  laws  this  equals  (x+a)  (x+d)  (x+c)  (x+d) 

ix+k)  (;r+/)=;tr«+i+  (P+l)x*'+  (P,+Pl)x''-^  + 

(P,  +  PJ)x^-^ P„l    (2.) 

Now  P+l=a  +  b  +  c+d+ +^+/ according  to  sec- 
tion three  of  the  foregoing  rules,  viz:  the  coefficient  of  the 
second  term  is  the  sum  of  the  second  terms  of  the  binomial 
factors,  and  there  are  now  /*+/ factors. 

P,  +  Pl=P^+  (a+b+c+d+    +k)l    equals 

the  sum  of  the  products  of  all  the  terms  a,  b,  c,  dy k, 

I  taken  two  at  a  time  by  the  foregoing  rule  and  demonstra- 
tion. 


Binomial  Formula.  3 

♦    p^^pj^p^Ji^{ab-\rac^r(id\bc-\-    a>^+^/^)  /  equals 

the  sum  of  the  products  of  all  the  terms  a^  b^  Cy  d .  .  .  .k^  I 
taken  three  at  a  time  by  the  rule. 

/>„ /=products  of  all  the  terms  a,  b,  c,  d, k,  I. 

The  law  of  the  exponents  in  (2)  is  the  same  as  in  (1). 

We  have  shown  by  former  demonstration  that  the  laws  are 
true  when  the  number  of  factors  n  is  two,  three,  four,  hence 
they  are  true  when  the  number  of  factors  ^^  is  five  and  so  on 
or  when  the  number  of  factors  is(;^  +  /). 

The  number  of  terms  in  /^  is.  obviously  n;   the  number  of 

terms   in    P^  is  equal  to  the  number  of  combinations  of  n 

1  •  ,  .  ,       .    n  (n — 1\      , 

thmgs,  taken  two  at  a  time,  that  is  — — ^;   the  number  of 

JL    X    ^ 

terms  in  P^  is  equal  to  the  combinations  of  n  things,  taken 

'  ^      •     n  (^—1)  (n—2)      ^ 

three  at  a  time,  that  is,  — ^ i-^ — — -^  and  so  on. 

*      1x2x3 

Now   suppose   a  ^=  b  =^  c  ^=    =   k;    then 

--  ^         n  hi — 1) 

P  =z  na,  P^  ■='—-^ — —    a"^  because  a  =^  b  ^=  c  =    

1x2 

7  J  J  7^  ^     (^—1)      (^ 2)  ,    ^ 

=^  k  and  so  on,  and  P^  = f -^ — - — -  a^  for 

^  1x2x3 

the  same  reason  and  so  on. 

Therefore  under  the  hypothesis   (1)    becomes    (^x   +  a)^ 
n  s  «   I    I     ^  (^  ~~  1)     2     ^    5  I   ^^  (^—1)  (^—2) 

^3  ^  «-3  _|_    _(.  na""-^  X  +  a^'^Z) 

We  now  have  formulated  and  developed  the  binomial 
formula.  The  second  member  of  (3)  is  called  the  expansion 
or  the  development  of  {x  +  ci)^. 

Corillary  (1) — The  sum  of  the  exponents  of  a  and  x  in 
any  term  of  the  expansion  of  {x  +  d)^  is  equal  to  «. 


4  Expansion  of  Quantities, 

To  obtain  the  expansion  of  {x  —  a)«  place  —  a  instead  of 
+  a  in  the  expansion  of  {x  +  a)^.  The  terms  which  con- 
tain the  odd  powers  of  —  a  will  be  negative  and  the  terms 
which   contain   the   even    powers   of  —  a  will  be  positive. 

Hence, 

n  (  n — 1  )     ^ 


(^x  —  a  )«  =:  x^  —  nax^~^    + 


1x2 


n{n  —  X)^n  —  ^)  n  {n  —  V){n —  ^)  {n  —  Z) 

1x2x3  "^  1x2x3x4 

a^  x"^-^ a"". 

Example. — Expand   {x  +  jk)^-     In  this   quantity  n  =  h, 

,,,      ,       6(5  —  1)(6— 2)(5  —  3)     ,,,, 
-^^  1x2x3x4  ^      J    -f 

5  (5-1)  (5-2)  (5-3)  (5-4)  ^,_, 


1x2x3x4  +  5 


:r5— 5  jj/5 


Simplifying  this  we  get  x^  +  bx^  y  +  lOx^  y""  +  10^*  j^' 
+  bxy^  +  y^. 

In  the  above  expansion,  the  following  order  may  be  ob- 
served. 

No.  of  terms= 
The  powers  of  jf= 
The  powers  of  j/= 
Coefficients^^: 

(^  +yy  =  xy  +  ^xy  +  lOxy  +  lOjr^j/^  +  bxy^  +y^x° 

It  will  be  observed  that  there  is  one  more  term  in  the 
expansion  than  the  number  of  the  given  powers,  viz:  5,  or, 
according  to  the  law,  the  number  of  terms  is  one  greater 
than  the  number  of  binomial  factors  in  the  first  member  of 


1 

2 

3 

4 

6 

6 

x^ 

X* 

x^ 

x'^ 

^' 

;c° 

r 

r 

r 

y' 

y 

f 

1 

5 

10 

10 

5 

1 

Expansion  of  Quantities.  5 

{x  +  yY,  and  the  coefficient  of  the  rth  term  from  the  begin- 
ning is  equal  to  the  coefficient  of  the  rth  term  from  the  end. 

The  exponent  of  x  decreases  by  one  for  each  succeeding 
term  of  the  second  member,  beginning  with  the  exponent  of 
the  required  power,  and  the  exponent  of  y  increases  by  one 
beginning  with  0. 

The  coefficient  of  the  first  term  is  one,  the  coefficient  of 
the  second  term  is  the  same  as  the  exponent  of  the  given 
power;  for  the  coefficients  of  the  following  terms,  multiply 
the  coefficient  of  the  preceding  term  by  the  exponent  of  x 
for  that  preceding  term  and  divide  by  the  number  of  pre- 
ceding terms,  as:  the  coefficient  of  the  third  term  is  the 
coefficient  of  the  second  term,  5,  multiplied  by  the  exponent 
of  X,  which  is  4  for  that  term,  this  equals  20,  and  divided  by 
the  number  of  preceding  terms,  which  are  two,  giving  10  as 
the  required  coefficient,  and  so  on. 

This  last  method  or  form  is  a  convenient  method  in  which 
to  place  the  expansion  of  any  binomial. 

Example.— Expand  (2^—3^/.       In    this    quantity  «=4. 

(2a  —  3^)^  =  (2a)4  —  4  (2^)3  U  +  ^/^~|^  (2a)^  (3^)^  — 

4(4-1)  (4-2)  4(4-1)  (4-2)  (4-3) 

1x2x3        ^^""^   ^^^^    +  1x2x3x4 

(2^)°  (3^)^  = 

16^4  _  96^3  jj  4.  2l6a^  b^—1VUb^  +  81^^ 
No.  of  terms==  12  3  4  5 


Powers  of  a^  (2^)*      (2a)3       {%ay      {2ay       (2^)° 

Powers  of  b^      —(3^)°  —(3^)'  —(3^)^  —(3^)3  —{Uy 
Coefficients=r  14  6  4  1 


Combining     1  (2^)^  (3^)°  —  4  (2a)3  (33)'  +  6  (2a)»  (33)^— 
4(2a)v(33)3  +  1  (2^)°  (3^)^  = 

16a^—  96^3  b  +  216a^  b^  —  216a  3^  +  31^4  ^s  before. 


6  Expansion  of  Quantities. 

c 

Observe  that  all  the  odd  powers  of  the  minus  quantity — Zb 

are  negative. 

Expand  (^a  -\-  b  -\-  c  -^  dy,  Let(«  +  ^)  =  w  and 
(^c  -\-  d^  ^==  n  , 

(m  +  n)^  =  m^  -{-  Zm^  n  +  Zm  n^  +  n^.     Substituting  for 
m  and  n  their  values,  {a-{-b)  and  {c-\-d),  this  gives,  (a  +  ^)^ 

a^  +  Za-b  ^Zab^  ^b'^  +  3(^j:^+  la  b -\- b^)  {c  ^r  ^)  +  3 

Example.— Expand  ( 46  )4. 

Let  45  =(  X  +  y)  or  40  =  ;»;,  and  5  ^=  jk 
{x  +  yy  ^x"^  +  4  x^y  +  6  jr^j/^  +  4  ^jv^  +y  sub- 
stituting values. 

(  40  y  +  4  X  403  X  5  +  6  X  40^  X  5^  +  4  X  40x  53  +  5^== 

2560000  + 1280000  +  240000  +  20000  +  625--4100625. 

Let  45 -=  (jr  —  j)  or  (50  —  5) , 

Then  (x  — yy  =  ;r^  —  Ax^y  +  ^x^'y''  —  Axy^  +  y^  = 
50*  —  4  X  503  X  5  +  6  X  50^x  5^—  4  x  50  x  5^  +  5*  == 

6250000  —  2500000  +  375000  +  25000  +  625  = 
4100625. 


N^^  Root  of  Quantities.  7 

CHAPTER  II. 
THE  N^^  ROOT  OF  QUANTITIES. 

We  have  already  shown  the  manner  of  developing  or 
expanding  a  quantity  by  the  Binomial  Theorem.  We  shall 
show  the  converse  of  this  process,  or  the  extraction  of  the 
nth  root  of  any  quantity  where  n  is   any  positive   integer. 

Any  given  quantity  may  be  considered  as  the  result  of  some 
expansion  or  the  continued  product  of  like  factors  of  the  re- 
quired root,  to  find  such  factors  is  the  result  of  the  following 
section.  The  nth  root  of  a  quantity  is  one  of  the  n  equal 
factors  of  that  quantity. 

Rule  for  the  extraction  of  the  »th  root  of  any 
quantity. 

First — Place  before  you,  or  keep  in  mind  the  general  form 
of  the  binomial  theorem ^  viz: 

n  (n  ^"^  1") 

(x+yy=x'^  +  nx^-\y  -\ ^ ^  x^-y^ y^ 

i.   X   ^ 

Second — Arrange  the  expression  according  to  the 
ascending  or  descending  powers  of  some  letter.  Extract 
the  nth  root  of  the  first  term;  the  result  will  be  the  first 
term  of  the  required  root,  or  the  equivalent  of  x  in  the 
general  equation.  Subtract  the  nth  power  of  the  first  term 
of  the  root  from  the  given  polynomial. 

Third  — Divide  the  first  term,  of  the  remainder  by  n  tiTnes 
the  (n — X)th  power  of  the  first  term,  of  the  root,  as  a  trial 
divisor;  the  quotient  will  be  the  second  term  of  the  root,  or 
the  equivalent  of  y  in  the  general  equation. 

Complete  the  divisor  by  adding  the  (n — 1)  remahiing 
terms  of  the  general  equation,  with  the  terms  of  the  root 
already  found  substituted  in  it,\to\  the  frial  divisor,  all 


8  N^^  Root  of  Quantities. 

divided  by  the  second  term  of  the  rooty  and  multiply  this  by 
the  second  term  of  the  root.  Subtract  this  product  from 
the  first  remainder  of  the  polynomial. 

Explanation — The  student  is  cautioned  to  remember  that  the 
n  referred  to  in  this  rule  is  the  n  of  the  required  root  or  in  other 
words  the  greatest  exponent  of  x  and  jv  in  the  general  formula 

Fourth — Take  n  times  the  (n — V)th  power  of  the  sum, 
of  the  first  and  second  terms  of  the  root  for  a  second  trial 
divisor.  Divide  the  first  term  of  the  second  remainder  by 
the  first  term  of  the  second  trial  divisor  ;  the  quotient  will 
be  the  third  term  of  the  root^  complete  the  divisor  as  before 
by  adding  the  (n — 1)  remaining  terms  of  the  general 
equation^  divided  by  the  third  term  of  the  root,  to  the  trial 
divisor  and  m^ultiply  this  sum  by  the  third  term  of  the 
root. 

Fifth — Proceed  in  this  manner  until  all  theTterms  of 
the  root  have  been  found. 

Example. — Extract  the  square  root  of  ^^  +  <5=  +  ^=' 

-^-lab^lac^Uc, 

«2  +  2ab  +  2^^  +  *lbc  ■\-b''-^c*\a-\-b^c 
a"" 


2ah  +  ^ac  +  2b c  ^  b""  ^  c  ""  =  X^^  remainder. 
2ab  +  b"" 

2ac  +  2bc  +c  *  =2nd        *« 

2ac  +  2bc  +c  ^ 

Trial  divisor  =  ^(/^ — 1)  power  of  a  =  2a        n  o-      -  ^^^ 
Complete  divisor  =2a  +  b 

the  second  term^       [(2^  +  ^)1^. 

Second  trial  divisor  n"^ — '-t).p&werjo{{a  +  b)  =  2  (a  +  b)  =  2a  +  2b 
Second  complete  divisor  2a  +  2b  +c 

"    thethirdterm"      ""'j^^a  +  26  +  c)  c 


N*^  Root  of  Quantities,  9 

Explanation— 

In  this  example  n^=^,   and  the  general  equation  with  2 
substituted  for  n  equals  x^  +  ^xy+y"". 

Three  distinct  terms  will  be   found  in  the  root,   because     h   / 
there  are  three  distinct  letters  in  the  given  example.  -^  ' 

Arrange  and  proceed  according  to  the  rule  (see  example) 

The  square  root  of  the  first  term  «^  is  <2,  or  the  x  of  the 
general  equation. 

The  second  power  of  a  is  a^.       Subtract   this  from   the 


given  polynomial,  leaving  the  first  remainder  as  shown.  ^ 


.^ 


I 


The  first  trial  divisor  is  n>{n — 1)  power  of  a,  or  2(2 — 1) 
power  of  Uy  which  gives  2a. 

Divide  the  first  term  of  the  remainder  by  2a;  this  gives  d 
the  second  term  of  the  root,  or  the  equivalent  y  of  the  general 
equation. 

Complete  the  divisor  by  adding  to  the  trial  divisor  the 
(^n — 1)  remaining  terms,  or  thej>/^of  the  general  equation, 
divided  by  y,  which  is  y  or  d,  this  gives  2a  +  5  as  a  complete 
divisor,  and  multiply  this  by  3,  which  gives  2ad+d'^.  Sub-  .^ 
tract  this  quantity  from  the  first  remainder,  leaving  2ac+2dc  -i 
+  ^^  as  a  second  remainder.  J* 

Now  treat  (a  +  d)  as  one  quantity  or  the  jtr  of  the  general  *^ 
equation,  and  according  to  rule  (Fourth)  take  n  or  2  times  ^ 
the  {n — 1)  power  of  (a  +  d)  or  2a  +  2d,  as  a  second  trial  | 
divisor.  X 

Divide  the  first  term  of  the  second  remainder,  2ac,  by  the  ^ 
first  term  of  the  second  trial  divisor   (2a);  this  gives  ^  the  •j  . 
third  term  of  the  root,  or  they  of  the  general   equation,  as  s$>* 
before.  -S 

Complete  the  divisor,  as  before,  by  adding  to  the    second    Ju 
trial  divisor  the  (n  —1)  remaining  terms  of  the  general  equation    ^ 
(which  in  the  square  root  is  always  one  term)  divided  by  the 
third  term  of  the  root,  this  gives  2a  +  2d+c,     Multiply  this 
quantity  by  the  third  term  c,   and  subtract  from  the    second 
remainder,  which  leaves  0. 

Therefore  (a  +  d+c)  is  the  desired  root  and  a,  d,  ^  are  the 
separate  terms  of  the  root. 


10 


N^^  Root  of  Quantities* 


+ 

+ 
+ 


+ 


CD 

+ 


be 

N^ 

c 

'% 

l-t 

o 

+ 

m 

1 

1 

V 

^ 

+ 

»»» 

-O 

*S 

« 

-«   + 


5 

Vj 

.o 

Ki 

vS 

« 

4) 

1—1 

•4J 

+ 

■M 

o 

cd 

<> 

i  ^ 

a    <) 
m    « 

+ 


+  + 


+  + 


W     tH    i-H     tH 


N*^  Root  of  Numbers,  11 

TO  FIND  THE  N'^^  ROOT  OF  A  NUMBER. 

The  square  of  a  number  may  contain  twice  as  many  figures 
as  the  given  number,  or  twice  as  many  less  one. 

The  cube  of  a  number  may  contain  three  times  as  many 
figures  as  the  given  number,  or  three  times  as;many  less  one 
or  two. 

The  rVCci  power  of  a  number  may  contain  n  times  as  many 
figures  as  the  given  number,  or  n  times  as  many  less  {n — 1) 
or  (;^— 2)  or  (/f— 3) 1. 

Hence,  if  we  separate  the  given  number  into  periods  of  n 
figures  each,  beginning  with  the  units,  we  shall  determine  the 
number  of  figures  in  the  required  root. 

Let  it  be  required  to  extract  the  square  root  of  50625,  a 
perfect  square.  50625  is  evidently  the  square  of  some 
quantity  of  three  periods  or  terms.  Let  a  represent  the 
value  of  the  first  figure  of  the  root,  b  the  second  figure,  and  c 
the  third  figure.  Hence,  {a-\-b-\-c)  is  a  factor  or  root,  and 
(a  +  b+cy  =  a^  +  2ab  +  b^  +  2ac+2bc  +  c^  equa\s  50625  or 
50000  +  600  +  25.  Now  follow  the  general  rule  already 
given,  remembering  that  ;2=2. 


5' 06' 25  I  200  +  20+5 

4  00  00 

1  06  25=  1st  TQma.\ndeY=2ab  +  b''  +  2ac+2bc+c' 
84  00  2ab+b'' 

22  25      2d  remainder     2ac+2bc+i^ 
22  25  2ac+2bc+c' 


Let  a=200  then  «»=: 200^=40000 

cr 

1st  trial  dWisor=n/(n — 1)  power  of  a=2a=400 

1st  complete  divisor=(2^+^)=  400  +  20=420 


12  Square  Root  of  Numbers, 

1st  complete  divisor  x  b  )  420x20=8400 
the  second  term  \ 

Second  trial  divisor=;2X(« — Vf^  power  of  (a  +  ^)= 

2  («+^)==2a  +  2^-=400+ 40=440 

2d  complete  divisor=2^  +  2^+<r=440  +  5=445 

^^  .u   '1-  ^  /'       ^^1  445x5==2226 
the  third  term  \ 

Therefore  the  square  root  of  50625  is  200  +  20  +  5=225. 

Explanation. 

Since  600  nor  25  cannot  be  part  of  the  square  of  the  hun- 
dreds a  must  be  the  greatest  multiple  of  the  hundreds  whose 
square  is  less  than  50000  ;  this  multiple  must  be  200.  Sub- 
tracting a^  or  40000  from  the  given  number,  we  have  the 
first  remainder,  10625,  this  equals  2ah-\-b^-\-1ac-\-^hc-^c^, 
We  already  have  the  term  a,  therefore,  if  we  double  the  term 
a  or  200  ^  7tf^n — 1)  power  of  the  first  term  of  the  root 
we  have  400,  the  trial  divisor. 

Divide  the  first  remainder  by  400  ;  this  gives  20,  the  sec- 
ond term,  b.  Complete  the  divisor  by  adding  this  term  to 
the  trial  divisor  and  multiply  by  the  second  term  ;  this  equals 
i^a^-b)  b  or  (400  +  20)  X  20=8400.  Subtract  this  from  the 
first  remainder  leaving  2225  or  ^ac-\-^bc-\-c'^.  We  now  have 
terms  a  and  b  eliminated  and  desire  term  c. 

Divide  the  second  remainder  by  rUin — 1)  power  of  the 
sum  of  the  first  and  second  terms  as  a  trial  divisor  or  2^  +  2^ 
=440  ;  this  gives  5  for  the  third  term  c.  Complete  the  divi- 
sor by  adding  this  term  to  the  trial  divisor  and  multiply  by 
the  third  term  ;  this  equals  (2a  +  2<^  +  ^:)  ^  or  ('440  +  5  )X  5= 
2225.  Subtract  this  product  from  2225  which  leaves  no  re- 
mainder ;  hence  200  +  20  +  5  =:  225  is  the  required  square 
root. 


Square  Root  of  Numbers,  13 

ROOT  CONTAINING  DECIMALS. 

In  the  extraction  of  a  root  containing  decimals,  care  must  be 
taken  to  divide  the  quantity  into  periods  from  the  unit  period 
as  before,  this,  of  course,  makes  the  division  of  the  periods 
from  the  left  to  the  right  beginning  with  the  decimal  point; 
and  in  a  quantity  containing  whole  numbers  and  decimals  the 
periods  should  be  spaced  from  the  right  to  the  left  for  the 
whole  numbers  and  from  the  left  to  the  right  for  the  deci- 
mals.    [See  following  example.] 

Extract  the  square  root  of  276.029056 

abode 
2^75.02^90^561  14  +  2  +  .5  +  .08  +  .004 
1  96 


79.02  90  56^1st  remainder 
60 


19.02  90  56=2d  remainder 
16.25 


2.77  90  56=3d  remainder 
2.64  64 


.13  26  56=4th  remainder 
.13  26  56 

Let  «=14  then  a2:^14^=196 


1st  trial  divisor=^^x(;^ — \y^  power  of  «=2x  14=28 

1st  complete  divisor^^lst  T,  Z>. +^=28  +  2=30 

1st  complete  divisor  x  b  \  qhv/  o  ac\ 

the  second  term         (  —^^  ^  ^'  —^^ 


14  Square  Root  of  Numbers, 

2d  trial  d:W\%ox=^n\{n — ly^'  power  of(a  +  (^)  =  2(«  +  ^)  = 
2  (14  +  2)=32 

2d  complete  divisor=2d  T.  Z>. +r=32  +  .5=:32.5 

2d  complete  divisor  x  ^  Ko  kv,    r_ifi  9^ 
the  third  term  \  ^^'^^  .5—16^ 

3d  trial  divisor  =  ;^  v(  n — 1  y^  power  of(a  +  <^  +  ^)  = 
2(a  +  ^+r)=2  (14  +  2  +  .5)=33 

3d  complete  divisor=3d  T.  Z>.+ ^==33 +  .08=^33. 08 

3d  complete  divisor  X  ^  )  qq  nc  v^  no       <^  aActi 
the  fourth  term         j  ^^'^^^  .08  =  2.6464 

4th  trial  divisor  «^« — \y^  power  of  («  +  ^  +  ^+^)= 

2  (a+iJ+/:+^)=2(14+2  +  .5  +  .08)=-33.16 

4th  complete  divisor=:4th   T,  Z>. +^=33. 16 +  .004=33. 164 

'^'u.^Sl^f  "l33.164x.004=.132656 

Therefore  the  square  root  of  275.029056=14  +  2 +  .5  +  . 08 
+  .004=16.584. 

Note  the  division  of  the  periods  from  the  decimal  point  to 
the  left  for  whole  numbers,  and  from  the  decimal  point  to  the 
right  for  the  decimals.  This  gives  two  periods  of  whole 
numbers  and  three  periods  for  decimals,  as  shown.  In  this 
example  proceed  as  before,  remembering  that  the  trial 
divisor  is  always  n  times  the  {n — \y^  power  of  the  sum  of 
the  terms  of  the  root  already  found.  The  complete  divisor 
is  simply  complying  with  the  general  equation  of  the  expansion 
of  (x  +  y^y  and  subtracting  this  expansion  for  a  new  re- 
mainder, if  there  be  one. 


Cube  Root  of  Numbers  15 

CUBE  ROOT. 

The  cube  root  of  a  number  is  one  of  the  three  equal 
factors  of  that  number. 

Example.— Extract  the  rube  root  of  2299968. 

a  b  c 
2'299'968  |100+30+2 
1000  000 


1299  968==  1st  remainder. 
1197000 


102  968=2d  remainder, 
102  968 


Let  a:-=100  then  ^3=1003=1000000 
1st  trial  divisor  =^7i{it — 1)^  power  of  a=3a*= 

3x100^=30000 
1st  complete  divisor  =lst  T,  D,+Sab+b''= 
30000 + 9000 + 900=39900 

'"^  1"et?ondtrm  '  \  =^9900  X  30  =  1.197000 
2d  trial  divisor  =^(:n—iy^  power  of  (a+b)=S(a+by 

Let  (a  +  b)=m.    Then  3(a+^)^=3w^=3x  130^=50700. 
2d  complete  divisor=2d  71  D.  +  Smc+c''= 
60700  +  780  +  4=51484 

Explanation. 

To  extract  the  cube  root  of  2299968  we  first  divide  the 
given  number  into  periods  of  three  figures  each,  beginning 
with  the  units;  this  gives  the  number  of  terms  in  the  required 


16  Cube  Root  of  Numbers, 

root  or  factor.  Hence,  2299968  may  be  considered  as  the 
cube  of  (a  +  ^  +  ^)  or  {a-\-b-\-cy  when  a^  b^  and  c,  or  their 
numerical  equivalents,  are  terms  of  the  complete  factor 
{a\-b^c). 

Since  299000  nor  968  cannot  be  part  of  the  cube  of  the 
hundreds,  the  term  a  must  be  the  greatest  multiple  of 
hundreds  whose  cube  is  less  than  2000000;  this  must  be  100. 
Subtracting  the  cube  of  a,  or  a}^=  1000000  from  the  given 
number,  we  have  the  first  remainder  1299968. 

We  already  have  the  first  term  a,  therefore  if  we  take  three 
times  the  square  of  the  first  term  or  7t\{n — \y^  power  of  the 
first  term  of  the  root  we  have  30000  as  a  trial  divisor. 
Divide  the  first  remainder  by  30000;  this  gives  30,  the 
second  term  of  the  root  or  b.  Complete  the  divisor  by  add- 
ing 3  times  the  product  of  the  first  and  second  terms,  plus 
the  square  of  the  second  term  already  found.  This  equals 
39900,  and  multiply  this  by  the  second  term  of  the  root  b  or 
30,  the  product  is  1197000.  Subtract  this  from  the  first 
remainder,  leaving  102968,  or  the  second  remainder. 

Divide  the  second  remainder  by  3  times  the  square  of  the 
sum  of  the  first  and  second  terms  as  a  second  trial  divisor, 
or  ii:f{n  —  \y^  power  of  (a  +  ^)  =  50700;  this  gives  2  as  the 
third  term.  Complete  the  divisor  as  before  by  adding  to 
the  trial  divisor  3  times  the  product  of  the  sum  of  the  first 
and  second  terms  by  the  third  term,  plus  the  square  of  the 
third  term;  this  gives  51484,  and  multiply  this  total  sum  by 
the  third  term,  giving  102968.  Subtract  this  product  from 
the  second  remainder,  which  leaves  no  remainder. 

Hence,  the  terms  of  the  root  are  100  +  30  +  2=132,  the 
complete  cube  root  of  the  number. 


Cube  Root  of  Numbers,  17 

Extract  the  cube  root  of  264732.956461. 

In  this  example  a  whole  number  and  a  decimal  is  given. 
Note  that  the  number  of  figures  in  each  place  are  three,  and 
they  are  pointed  off  from  the  decimal  point  to  the  left  and 
right  for  the  whole  number  and  decimal  respectively. 

abed 
264732.956^461  |  60  +  4  +  .2  +  .01 
216  QQO 

48  732.956  461=lst  remainder 
46  144 

2  588.956  461=:2d  remainder 
2  465.288 


123.668  461=-=3d  remainder 
123.668  461 

Let  a^m  then  a?=m^^VLmm 

1st  trial  divisor  7i^{n — Vf^  power  of  a=3  a^= 

3x60^==:10800 

1st  complete  divisor=lst  T.  B.+Sab  +  b''= 

10800+720+16=11536 

1st  complete  divisor x^  )       hko/jv.  a       ag-iaa 
the  second  term      }  =11536x4  =  46144 

Let  (a  +  b)=m 
2d  trial  divisor=;ey(  n  — - 1  )  ^^  power  of  m  =  Sm^  == 

3x64^=12288 
2d  complete  divisor=2d  T.D.  +  Smc+e''= 
12288  +  38.4+.04=12326.44 

'"  Te'ihfrdtrm  .2=2465.288 

lyet  (a  +  b  +  cy-==n 
3d  trial  divisor:=;2X^ — 1)^^  power  of  n=Sn^ 
3x64.2^=12364.92 


18  Cube  and  §*^  Root  of  Numbers. 

3d  complete  divisor==:3d  T,  D.  +  Snd+d^ 
12364.92  +  1. 926  +  . 0001==12366. 8461 

^"^Te^iou'^^^^^^^^^  I  12366.8461x.01=123.668461 

Therefore  the  cube  root  of  264732.956461=60  +  4  +  . 2  + 
.01=64.21. 


Extract  the  fifth  root  of  122298103125. 

a        b      c 
12' 2298r  03125  |  100  +  60  +  5 
1  00000  00000 


11  22981  03125=lst  remainder 
9  48576  00000 


1  74405  03125=2d  remainder 
1  74405  03125 

Let  ^=100  then  aS^ioo^^  10000000000 
1st  trial  divisor=  ny(^n — 1)  ^'^  power  of  a=6a'^^= 

5x100^=500000000 
1st  complete  divisor=lst  T.D.+5(2a^b  +  2a^b''  +  ab^)  +  b^==^ 

500000000  +  600000000  +  360000000  +  108000000  + 

12960000=1580960000 

^''  tTeTcTnd'ttm  ""  ^  |  1580960000X  60=94857600000 

Let  (a  +  b)=m 
2d  trial  divisor=;2)(^(;2 — ly^-  power  of  m=6m^= 

5x160^=3276800000   . 
2d  complete  divisor==lst  T.n,+5(2m^c  +  2m^c''  +  mc^)  +  c^= 

3276800000  +  204800000  +  6400000  +  100000  +  625= 

3488100625 

^^  t^ffird  ^^m'""'  I  3488100625  X5.=17440503125 


^^^  Root  of  Numbers.  19 

Explanation.  . 

To  extract  the  fifth  root  of  122298103125.  Divide  the 
given  number  into  periods  of  five  figures  each,  beginning 
with  the  units;  this  gives  three  terms  or  figures  in  the  required 
root  or  factor.  Hence,  122298103125  may  be  considered  as 
the  fifth  power  of  {a-\-b-\'C)  when  a,  ^,  and  c^  or  their  numer- 
ical equivalents,  are  terms  of  the  complete  factor  (a  +  ^  +  ^.) 

The  first  term  a  of  the  fifth  root  of  the  given  number 
122298103125  must  evidently  be  100,  since  it  is  the  greatest 
multiple  of  hundreds  whose  fifth  power  is  less  than  120000- 
000000.  Subtracting  the  fifth  power  of  the  first  term,  or 
10000000000  from  the  given  number,  we  have  the  first 
remainder. 

Divide  the  first  remainder  by  the  first  trial  divisor,  or  5x 

iOO'*  or  500000000,  we  obtain  60  as  the  second  term  of  the 

root.     Complete  the  divisor  by  adding  to  the  first  term  the 

.  terms  indicated  by  the  general  formula,  and  multiply  by  the 

second  term.     Subtract  and  proceed  as  before. 

Let  us  analyze  this  example  and  note  its  relation  to  the 
general  binomial  formula. 

Let  {a  +  b)  =  m.  Therefore  a  +  b  +  c^  {m  +  c).  Expand  (w  +  <:) 5. 
This  gives  7n^ -\-hm^c +l^m^c^ +\(im''c^ +bmc^  -\- c^.  Substitute 
given  values  and  expand  further,  {a  +  b)^ +  b(a^-b)^c\-lQ{a-\-b)^ 
c^-^lQ{a  +  byc^  +  b{a-\-b)c^^-c^ 

+  5(a4  +  4a3<^  +  6«='<^"  +  4a/^3  +  ^4y=  5^4^ 

+  6(a  +  b)c^  =  bmc^ 

a  equals  100,  or  the  greatest  multiple  of  hundreds  whose 
fifth  power  is  less  than  120000000000.  ^^-^  10000000000. 
Subtracting  this  from  the  number  122298103125  gives  the 
first  remainder    112298103125,   or    the    remaining    portion 


20  5^^  Root  of  Numbers. 

5a^<^  +  10«3^^  +  10a^^H- +<:^     The  trial  divisor 

is  n\(^n — ly^  power  of  a  equals  ba"^  or  5  X  (100)'^= 
600000000.  Divide  the  first  remainder  by  this  quantity. 
500000000,  which  gives  the  second  term  ^=60  of  our  root, 
We  now  have  the  [a  +  b)  terms  of  our  root.  Complete  the 
divisor  by  adding  to  the  trial  divisor  the  remaining  terms  of 
(a-\-by  divided  by  b,  or  ba^^-\Oa^b-\-lOa^b^-\-bab^  +  b^= 
1580960000,  and  multiply  this  by  /5==60.  This  gives  5a^^  + 
lQa^^  +  10a^b^  +  bab^-\-b\  which  equals  94857600000,  the 
complete  fifth  power  of  the  first  two  terms  of  the  root,  or  160. 
«  The  second  trial  divisor  is  evidently -ttfe  Tp^n — ly^  power 
of  {a  +  b)  or  5  (  «^  +  4.a^b  +  ^a^b^  +  4:a  b^  +  b^)  or  letting 
(a  +  ^)=^«  as  before  to  avoid  lengthy  terms  the  trial  divisor 
=bm\  (^-l-^)-=160.  5;;^^=:160^--3276800000.  Divide 
the  second  remainder  by  this  quantity,  this  gives  ^  or  6  the 
third  term  of  the  root.  Complete  the  divisor  as  before  by 
adding  lOm^c+lOm'^c'^  +  bmc^+c^  to  the  trial  divisor.  The 
complete  divisor  is  now  5m^  + 10 m^c+ 10 m''c''  +  5mc^  +  c*. 
Substituting  the  numerical  values  for  m  and  c  in  this  com- 
plete divisor  gives  3488100625,  and  multiply  this  quantity  by 
c=5  the  third  term,  this  gives  17440503125.  Subtract  this 
product  from  the  second  remainder  which  leaves  no  remainder. 
Hence,  the  complete  root  is  the  sum  of  the  terms  a  +  b  +  c  or 
100  +  60  +  5=165  as  the  fifth  root  of  the  given  quantity. 


Theory  of  Equations,  21 

CHAPTER  III. 
THEORY  OF    EQUATIONS. 

It  has  been  thought  advisable  to  prelude  the  method  for 
the  solution  of  equations  by  some  notes  on  their  general 
theory  for  the  guidance  and  simplicity  in  solution  :  as  the 
following  examples  will  show. 

Every  equation  of  the  7^th  degree,  containing  one  un- 
known quantity  may  be  written  under  the  form,  x^ ■\- ax^'^ -\- 

bx^  2 -\-kx-\-L^=^,      This    equation  is  called   the 

general  equation  of  the  n\.\i  degree.  The  term  Z,  or  the  ab- 
solute or  independent  term,  may  be  considered  the  co- 
efficient of  ;t:°. 

A  function  of  a  quantity  is  an  expression  containing  that 
quantity  ;  thus  ax^-\-bx  is  a  function  of  x. 

The  symbolsy  (:v),  F  {pc)^  are  sometimes  used  for  brevity 
to  denote  the  function  of  x. 

Any  quantity,  which  substituted  for  x  in  F  {x)  and  causes 
F  {pc)  to  vanish  is  a  root  of  the  equation/"  (jc)=0;  or  in  other 
words,  a  root  of  an  equation  is  an  expression,  which,  when 
substituted  for  the  unknown  quantity  will  satisfy  the  equa- 
tion. 

If  F  {pc)  vanishes  when  x=r,  the  function  is  divisible  by 
{pc  —  r),  \i  F  {x)  is  divisible  by  (  x —  r),  then  r  is  a  root 
of  the  equation  /  (^)  =  0.  When  F  (^)  is  of  the  ;^th  de- 
gree, it  has  n  roots  and  no  more. 

Theorem. — To  find  the  relation  between  the  coefficients 
of  y  {x)  and  the  roots  of  the  equation  y  {x)  =  0. 

Suppose  the  terms  of  /  {x)  to  be  arranged  according  to 
the  descending  powers  of  x  and  that  the  coefficient  of  the 
first  term  is  1;  then  : 

1.  The  coefficient  of  the  second  term  with  its  sign 
changed  is  equal  to  the  sum  of  the  roots. 


^22  Theory  of  Equations. 

2   5^       2.      The  coefficient  of  the  third  term  is  equal  to  the  sum 
3^  S'     ^f  the  products  of  the  roots,  taken  two  and  two. 

^  \  3.  The  coefficient  of  the  fourth  term  with  its  sign  changed 
I  "tf^  is  equal  to  the  sum  of  the  products  of  the  roots ^  taken  three 
I   ^     and  three  and  so  on, 

<    »         ^.     If  the  degree  of  the  equation  is  even^  the  absolute 

"^  g     term  is  equal  to  the  product  of  all  the  roots.     If  the  degree 

^\    of  the  equation  is  odd^  the  absolute  term  with  its  sign  changed 

^'    is  equal  to  the  products  of  all  the  roots, 

^        Cop.  1.     If  the  roots  of  y  (jc)=0  are  all  negative,  each 

term  oi  f  {oc)  is  positive. 


I 


I         Cop.  2.     If  the  roots  of/(^)=Oare  all  positive,  the 
v^  ^^^^  signs  of  the  terms  of  y"(^)  will  be  alternately  +  and  — . 

S"   ^       Cor.  3.     If  the  second  term  of  /  {oc)  does  not  appear,  the 
^    ^    sum  of  the  roots  of  the  equation/ (^)  =  0  is  equal  to   zero. 

gS  I        Cop.  4.     If/(^)  has  no  absolute  term,  at  least  one  of  the 

^  I     roots  of  y  (^)  =  0  is  zero.     Thus,  one  root  of  the  equation 

..5     x^ — 2jr' +  3^  =  0  is  zero. 
Mr?  5 
3    I         Cop.  5.     The  absolute  term  of  /  {x)  is  divisible  by  each 


J  S  i^'  root  of  the  equation  /  {x)  =  0. 


5*        Cop.  6.      Let  a,  b,  Cy  d /  denote  the  roots  of  the 

^  <^  equation  ^«  +  Ax''-^+  Bx""-^ +  K;t;  +  Z  =  0; 

then -^  =  a  +  ^+^  +  ^+ +/ 

and      B  =^ab  -\-  ac  ■\- -^bd  ■\-  be  ^r 

whence   A^—2B  =  a^  +b^  +  c^  +d' +  /"; 

5^  that  is,  .4= -2^  is  equal  to  the  sum  of  the  squares  of  the 
roots  of  the  proposed  equation.  Hence  if  A"" — 2B  is  nega- 
tive, the  roots  of  the  equation  cannot  be  all  real.''^Thus,  the 
roots  of  the  equation  x^—4:X^  +  22x^ — 25ji;— 42==0  are  not 
all  real    for  ( — 4)=^— 2x  22  is  negative. 


Theory  of  Equations,  23 

An  equation  whose  coefficients  are  integers,  that  of  its 
first  term  being  unity,  cannot  have  a  root  which  is  a  rational 
fraction. 

Let  the  equation  be 

^n  J^Ax""-^  +^;r«-2 ^rKx-\-L={^ (1) 

in  which  the  coefficients  A,  B, A^,  Z,  are  supposed  to 

be  integers. 

Suppose,  if  possible  that  (1)  has  a  rational  fraction  root 

which  in  its  lowest  terms  is  expressed  by  -.     Substituting - 

for  X  in  (1),  and  multiplying  the  resulting  equation  by  3^— i, 
we  obtain 

^^+Aa''-^+Ba^-'^b+ +  Kab''-'^  +  Lb''-^=  0 (2); 

whence, 

I  ^=  —{^Aa^-'^-^Ba^'-H ^  Kab^-'^  ^- Lb^-^) (3). 

The  second  member  of  (3)  is  an  integer,  and  its  first  mem- 
ber is  an  irredi\cible  fraction.     Hence  -  cannot  be  a  root  of 

b 

the  proposed  equation. 


DESCARTES^    RULE    OF    SIGNS. 

A  complete  equation  is  one  in  which  no  power  of  x  is 
wanting. 

In  any  series  of  quantities  a  pair  of  consecutive  like 
signs  is  called  a  permanence  of  signs  and  a  pair  of  con- 
secutive unlike  signs  is  called  a  variation  of  signs.  Thus  in 
the  expression  x"^ — x^ — hx^  +  \x^  +  ^x"^ — Zx^ — 6^  +  7—0 
there  are  three  permanences  and  four  variations. 


24  Theory  of  Equations. 

Theorem  of  Descartes*. — The  number  of  real  positive 
roots  of  the  equation  f  (^x)^=^  cannot  exceed  the  number  of 
variations  in  the  signs  of  its  terms;  and  if  the  equation  f  ix) 
=0  is  complete  the  number  of  real  negative  roots  cannot 
exceed  the  number  of  permanences  in  the  signs  of  its  terms. 

A  complete  equation  whose  signs  are  all  positive  can -have 
no  positive  real  root,  for  there  is  no  variation  of  signs. 
When  the  signs  are  alternately  positive  and  negative,  there 
is  no  negative  real  root  for  there  is  no  permanence  of 
signs. 

If  y  (jr)=0  is  of  an  odd  degree,  it  has  at  least  one  real 
root ;  if  of  an  even  degree  all  the  roots  may  be  imaginary. 


TRANSFORMATIONS    OF    EQUATIONS. 

To  transform  an  equation  containing  fractional  coeffi- 
cients into  another  in  which  the  coefficients  shall  be  integers, 
that  of  the  first  term  being  unity. 

Let  the  proposed  equation  be 

x^  +Ax''-'^  +  Bx''-^.  .....  +J^x  +  L=0. (1), 

in  which  some  or  all  the  coefficients  Ay  B^  C, A"  are 

supposed  to  be  fractional. 

Assume  \.h2X  y=kx,  or  x=-^.     Substituting  -^  for  x  in  (1) 

and  multiplying  the  resulting  equation  by  y^«,  we  have 

y^  +  Aky^'-^  +  Bky-'^ +  Kk'^-^y  +  Lk'' =-0 (2). 

Now  since  k  is  arbitrary,  we  may  give  it  such  a  value  as 

will   make   the    coefficients   Ak,    Bk', Kk^-^,    Lk'' 

integers. 


Solution  of  Equations  of  Higher  Degrees,         25 


CHAPTER  IV. 

SOLUTIONS  OF  EQUATIONS  OF  HIGHER 
DEGREES. 

In  the  previous  chapters  we  have  reviewed  a  comprehensive 
and  detailed  demonstration  of  the  binomial  formula,  pro- 
cesses of  involution  and  evolution,  and  the  general  theory  of 
equations,  for  the  purpose  of  our  present  chapter,  namely, 
the  solution  of  algebraic  equations  of  higher  degrees. 

Let  ^jtr^+^jr'^-i+Cr'^-H^^''-^ +A"=0 

be  a  general  equation  of  the  ^^th  degree,  where  x  is  the 
unknown  quantity,  and  A,  B,  C,  D^  etc.,  coefficients  of  x, 
and  K,  the  absolute  term,  which  does  not  contain  x. 

In  the  above  general  equation  the  quantity  sought  is  the  ^j 
value  of  X  that  will  satisfy  the  above  equation. 

The  above  equation  implies  that  the  absolute  term  is  A 
times  the  n\.\\  power  of  jt:,  +  ^  times  the  {n — 1)'^  power  of 
x,+  C  times  the  (n — 2)  power  of  x,  +  etc. ;  hence,  if  by  some  ^ 
means  we  can  simultaneously  extract  the  nth  root,  the  ^ 
(;^ — ly/i  root,  the  (n — 2)^^  root,  of  the  various  terms  of  the  *^ 
general  equation,  divided  by  the  various  coefficients  A,  B,  C,  | 
etc.,  we  shall  obtain  the  desired  result,  and  produce  such  ^ 
factor  or  factors  that  will  satisfy  the  above  equation.-^  ^^ 

Hence,  to  solve  any  equation  of  one  unknown  quantity, 
simultaneously  extract  the  various  roots,  indicated  by  the  ^ 
various  exponents,  of  the  terms  of  the  given  equation,  divided  "^ 
by  the  various  coefficients.  ^ 

RULE  FOR  THE  SOLUTION  OF  EQUATIONS  b 

OF  THE  A^H  DEGREE.  | 

Ax''+Bx''-'^+Cx''-'^+Bx''-^+ +J^=0        (1) 

In  the  above  equation  let  x=(a  +  d)  where  a  and  d  are 
terms  of  the  desired  root  of  the  equation;  that  is,  if  (a  +  d) 
were  substituted,  it  would  satisfy  the  equation. 


I 


V 


.26  Solution  of  Equations, 

(1).     Substitute  this  value  into  the  given  equation  and 
^^  5<^  expand  according  to  the  binomial  theorem. 

V  ^  A  {a-\-bY=    1st  term 

^'  ^'  +^  («  +  ^)«-i=:2d  term  ^        .       ,^, 

^  ^  +C  (a  +  ^)-2__^3d  term  ^  ^^"^^^^"  ^2)- 

2    5  +A^:=0  Absolute  term. 


?        This  expanded,  equals: 
^'  s  '       A{a^-\-na^-HV^^^~^\''-H'' .  , . .  +^'^)=lst  term^ 

rs'^  b''-^)=  2d  term  I 

v2^^  +C(a'^-2  +  (^— 2)a«-3^+ etc.)-=  3d  term  I 

+A"=0  Absolute  term/ 


(3). 


(2).  Separate  and  arrange  the  terms  of  (3)  so  that  all  the 
highest  powers  of  a  in  each  consecutive  term,  beginning 
with  the  highest  term,  are  consecutive,  as:  Aa^-\-BQk;^—^-\- 
Gj«-2+Z?^«-3+etc.    ^1 

(3).  Arrange  the  terms  so  that  all  the  n{n — Xf^  powers 
of  the  expanded  consecutive  terms  in  (3)  are  consecutive, 
as:  (.4;^a'^-l  +  i5(«— l)a''-2+C(«— 2>'^-H 
etc.)  b.  These  quantities  are  evidently  the  trial  divisors  for 
the  extraction  of  the  various  roots  indicated  by  their  respec- 
tive exponents  of  the  various  powers,      pf, 

(4).  Arrange  the  remaining  terms  so  that  they  are  placed 
according  to  their  respective,  positions  in  the  general  equa- 
tion, beginning  with  the  highest  remaining  term  of  a. 

The  equation  may  now  be  arranged  thus: 
(^^«  +  i5a«-i+Ca«-2+-^«''-H )=lst  term 

+ !)^2d  term 


Solution  of  Equations.  27 

+  (^^^«''-^+  + B^^^^^^^a-'-H^  +  + 

^(^_l)(^_2)(^-3)^^  „^_^  +  +)]*=3d  term 

L    + K=0=         Absolute  term.         =Equation  (4).      <i 

^        (6).     Substitute  some  trial  value  for  a  in  the  first  term  of     « 
(3)  whose  value  is  equal  to,  or  less,  than  K  preferable,  if         ^ 

(6).  Subtract  this  value  from  K,  which  gives  a  first  j 
remainder.  This  evidently  equals  the  remaining  terms  of  J^ 
equation  (4).  ^^ 

(7).     Divide  the  first  term  of  the   remainder  by  the  trial 
divisor,  or  the  second  term  of  equation  (4);  the  quotient  is   ^ 
the  second  term  of  the  roo^VComplete  the  divisor  by  add-    n^ 
ing   the  remaining  term  of  equation  (4),  and  multiply  this    ^^ 
sum  by  the  second  term  of  the  root.     Subtract  this  product 
from  the  first  remainder. 

(8).  Take  the  sum  of  the  first  and  second  terms  of  the 
root  for  the  value  of  «',  and  substitute  it  into  the  second  term 
of  equation  (4)  for  a  second  trial  divisor.  Divide  the  first 
term  of  the  second  remainder  by  the  second  trial  divisor; 
the  quotient  is  the  third  term  of  the  root.  Complete  the 
divisor  by  adding  the  remaining  terms  of  equation  (4)  and 
multiply  this  by  the  third  term  of  the  root.  Subtract  this 
product  from  the  second  remainder  for  a  third  remainder. 

(9).  Proceed  in  this  manner  until  all  the  terms  of  the 
root  have  been  found. 

(Note. — The  student  may  fiad  some  difficulty  in  finding  a  trial  ^  vS 
divisor  as  stated  in  (5),  whose  value  substituted  in  the  first  term  "^  ^ 
of  equation  (4)  will  be  equal  to  or  less  than  K,     When  the  value    ."^ 
equals  A' the  equation  is  evidently  solved.     If  the  term  substituted 
is  greater  than  K,  this  will  only  give  a  minus  quantity  as  a  first 
remainder,  but  no  way  impair  the  solution  of  the  given  equation. 
By  following  the  solution  of  the  simpler  numerical  equations  that 
follow,  the  process  will  be  clearly  shown.)  WU 
iiU^im  tolAi  Ho  eauijy&n^  ^C^moAM  Un^a^i^^CM  >5^^  ^  1\mA^C4XAC  ccu^ 


3 


28  Solution  of  Equations. 

Example — 

Solve  the  equation  ^^  +  jtr=650 

x^+x — 650-r°=:0.  In  this  equation  we  have  one  per- 
manence and  one  variation,  hence  it  contains  one  positive 
and  one  negative  root  according  to  Descartes'  rule. 

From  Cor.  6,  page  (22),  A^=l''=^the  sum  of  the  squares 
-^     of  the  coefficients,  or  the  coefficient  of  the  second  term. 

The    coefficient    of    the    third    term,    B  x°  ^=   —  650. 
"^  T    Therefore— 2^=1300. 
^    ^  A"" — 2^=1  +  1300   a    positive   quantity;    hence,   all    the 

f*      roots  are  real.    -H- 
%^  I  x^  +  x=Q50 

P    1^  ad 

^  I  '  650  I  20  +  5 

^    ^  420 

I    ^  230  [(2a  +  l)  +  ^]^=lst  remainder, 

r*:^  230 

ti  f : 

I         Let  ^=20,  then  a^  +  a^20"  + 20^420 

I  J^    1st  trial  divisor=2a  +  l=r2x  20 -t- 1=41 

V   f    1st  complete  divisor=3lst  T.  Z>. +^=41  +  5^:46 

^   "V^  1st  complete  divisor  x  ^  )       ,^     ^     ^^^ 
5-  ^  .u  A  .  =-46x5=230 

*     n  the  second  term        ) 

.j» 

^1  20+5=25 


I 


jr=25 
^—25=0 
x^+x—650  \x—2b 


x^—25x  X  +  2Q 

26jr— 650 
26x— 650 


Therefore  the  roots   are   25  and  — 26,  and  the  factors  of 
;»;=^+-;f— 650=0  are  (jr— 25)  and  (:r  +  26). 


Solution  of  Equations,  29 

Explanation. 

There  is  involved  in  this  equation  a  second  and  a  first 
power  of  x\  hence,  to  determine  the  roots  or  factors  of  x  we 
must  evidently  simultaneously  extract  the  first  and  second 
roots  of  the  absolute  term  divided  by  the  coefficients  of  the 
several  powers  of  .r. 

Apply  the  rule  given  on  page  (25)  for  the  solution  of 
equations. 

Let  X  equal  some  quantity  of  two  or  more  terms  :  (By 
terms  are  meant  quantities  whose  algebraic  sum  is  equal  to 
the  root  of  the  given  unknown  quantity.)  as  {a-\-b)  or  (^a-\-b 
+  ^).  Now  substitute  this  value  in  the  given  equation  ; 
therefore  jr^+;»;—650=0=(a  +  ^)"  (1st  term)  +  (a  +  ^)  (2nd 
term)— 650=0.  (1).  This  expanded  equals  (a»  +  2a^  + 
^^)  +  («  +  ^)— 650=0.     (2). 

Separate  and  arrange  according  to  (2)  of  the"]^rule  where 
a'^-\-a  are  the  highest  powers  of  a  in  the  above  consecutive 
terms.  (2a  + 1)  are  respectively  the  iUji — 1)  powers  of  d^  and 
a,  the  first  terms  of  the  expanded  consecutive  terms. 

The  equation  may  now  be  written  in  this  form  to  comply 
with  the  rule.    (a^+^)  +  [(2^  +  l)  +  ^]^=650as  per  (4).    (3). 

Now  assume  a  equals  20  as  the  first  term  of  the  root  and 
substitute  this  in  the  first  term  of  (3)  according  to  (5)  of  the 
rule.  ^='  +  ^=20*  + 20=420.  Subtract  this  value  from  650 
which  gives  230  as  a  first  remainder  ;  this  (230)  evidently 
equals  [(2a  +  l)  +  <5]<^  for  a  is  already  known. 

Now  divide  the  first  remainder  by  the  trial  divisor  (2a +  1) 
or  2x20  +  1=41.  This  gives  5,  the  second  term  of  the 
root.  Complete  the  divisor  by  adding  the  remaining  terms 
of  (3)  or  the  b=^h  just  found,  this  gives  2a  +  l  +  ^=41  +  5= 
46  as  a  complete  divisor  and  multiply  by  <^  or  5  ;  46x5= 
230.  Subtract  this  product  from  the  first  remainder,  leaving 
no  remainder;  hence  the  terms  are  a=:20,  b^=h  and  (a  +  ^)= 
25  is  the  required  root.  Therefore  x=^h  and  x — 25  is  one 
factor. 


30  Solution  of  Equations, 

There  are  as  many  roots  as  the  power  of  the  highest 
exponent  of  the  unknown  quantity  ;  hence  two  roots  in  this 
equation.     Page  21. 

If  x—2b  is  a  factor  of  jtr^  +  jr— 650^0  ;  x^'  +  x—^b^  is 
divisible  by  ji; — 25.  Dividing  as  shown  gives  jt:+26  as  the 
other  factor. 

Therefore  ;t;=|_26^"^('^~^^)(-^+^^)^-^'  +  -^~^^^""^- 

Let  us  suppose  that  in  the  same  equation  jt:^+;r=650;  we 
had  three  terms  in  our  desired  factor  or  root,  and  solve  under 
these  conditions;  then  x={a  +  b  +  c)  and  x^+x  would  equal 
to  (a  +  d+cy+(a  +  d+c).  Expanding  this  equals  a^  +  2a<^+ 
d'  +  2ac+2dc+c^+a  +  d+c. 

Separating  the  terms  according  to  our  rule  equals  a^  +  a  + 
[(2a+l)  +  ^]^+[(2a  +  2^+l)+^>. 

a      b      c 
660  I  10  +  10  +  5 

110 

540  [(2a  +  l)  +  ^]^+[(2«  +  2^+l)  +  <;>=lst  remainder. 
310  [(2a  +  l)  +  ^]^ 

230  [(2a+ 2^+ l)+^]r=2d  remainder. 
230  (a+^+0=25 

^=25  as  before. 
Let  g^lO.  then  a'+a=10'  +  10=110 
1st  trial  divisor=2«  +  l=2x  10+1=21 
1st  complete  divisor=(2^  +  l)  +  <^=21  + 10=31 

1st  complete  divisor x  b  )  o^  v,  -iq o-jq 

the  second  term         J 

2d  trial  divisor=:(2a  +  2^+l)=20  +  20  +  l=41 

2d  complete  divisor:=2d  T,  Z?. +c=41+5=46 

2d  complete  divisor  x  r )  . g     - qqq 

the  third  term         J 


Solution  of  Equations,  31 

Thus  any  number  of  terms  may  be  used,  and  a  oi  x  may 
be  represented  by  any  numerical  quantity  if  the  requisite 
signs  be  given  to  them  as  we  proceed;  for  example,  let  us 
assume  ;r=30,  which  is  greater  than  the  desired  factor,  as 
our  solution  has  shown. 

a      b 
650  I     30—5 


Let  ^=30,  then  a2  +  a=30^  +  30=  930 

1st  trial  divisor=2a  + 1=2x30  +  1=61  —280 

1st  complete  divisor=(2a  +  l  +  ^)=  — 280 

61  +  (— 5)=61— 5=66 

1st  complete  divisor  x  b  ) ^n^      k oon 

the  second  term  \  — i)bX  — &— J8U 

Example. 

Solve  the  equation  3:^^+4^=224. 

In  this  equation   x"^  and   x    have  coefficients   3   and    4 
respectively,  therefore  224   is   3    times  the  square  of  some    ^ 
quantity  plus  4  times  the  first  power  of   the  same  quantity. 
To  solve  this  equation  means  to  obtain  the  sum  of  the  ex-      J^ 
.      traction  of  the  square  root  plus  the  first  root  divided  by  their     "^ 
\     respective  coefficients.  ^ 

3jr'  +  4:t: — 224=0.  In  this  equation,  we  have  one  per-  ^ 
manence  and  one  variation  :  hence  one  negative  root  and  one  ^ 
positive  root.  J 

"^   .4^— 2^=4— (—448)=16  + 448=464,  therefore   all   the^A- 
roots  are  real.     Cor.  (6)  page  (22).  '^^'v 

Let  x={a^-b),  therefore  3^^  +  4:i:— 224=0.       Z{a-\-by  ** 

+  4(a  +  <^)— 224=0.     Expanded=3(«^  +  2a^+^^)  +  4(fl  +  ^)  J- 

=224.     Arranged  and  separated  according  to  the  general  ^ 

rule,  this  equation  equals  (3a^  +  4a)  +  [(6a  +  4)  +  3^>=224,  "^ 

where  (3^^  +  \d)  correspond  respectfully  to  the  Aa^  and  Ba^-^  "^ 


32  Solution  of  Equations, 

of  the  general  equation  :  and  (6a +  4)  are  respectfully  the 
n{n — \y^  powers  of  Za^  and  4a,  or  the  first  trial  divisors  of 
the  given  equation. 

a      b     c 
224  I  3  +  4  +  1 

39 
185  1st  remainder. 
136 

49  2d  remainder. 

49 

Leta=3,  then  3^^  +  4^:^=8x3^  +  4x3^27  +  12^39 
1st  trial  divisor=6a  +  4=6x  3  +  4-=:22 
1st  complete  divisor=lst  T.  Z). +3^=22  +  12=34 

'1      "     H."     "'1=34X4=136 
the  second  term  ) 

Let  (a  +  d)=m 
2d  trial  divisor=6 w  +  4=6  X  7  +  4=46 
2d  complete  divisor=:2d  T.  Z>. +  3^=46  +  3=49 

^^        ''  ''     ^M  =49X1=49 

the  third  term 

3  +  4  +  1=8 

^=8 
;r— 8=0 

3x^+  4:r— 224  I  x— 8 


3jr'^— 24^ 3ji:+28 

28:r— 224 
28^—224 

BX^  +  4:X=224:. 

28 
Therefore  3;i;+28=0.     3.^=— 28.     x= 3=—^/^- 


Solution  of  Equations.  33 

Explanation. 

Referring  to  the  example  on  page  (31),  assume  that  a=3 
in  the  equation,  then  (3^''  +  4a)=39.  Subtract  this  from 
the  given  value  224,  this  gives  the  first  remainder  185. 

185  evidently  equals  [  (6a  +  4)  +  3^]<^.  The  value  of  ^  has 
been  found;  b,  the  second  term,  is  now  desired.  (6^  +  4)  is 
our  trial  divisor.  Substitute  the  value  of  (3^=3  in  the  trial 
divisor,  this  equals  22,  and  divide  the  first  remainder,  185, 
by  this  quantity.  Let  the  quotient  be  4=<5,  or  the  second 
term  of  the  root.  Complete  the  divisor  by  adding  Zb  or  12, 
and  multiply  by  b  or  4.  [  (6^  +  4)  +  3/^]^=[  (6x  3  +  4)  + 
12]  X  4=136. 

Subtract  this  product  from  the  first  remainder,  leaving  49 
as  a  second  remainder;  hence,  another  term  of  the  root  is 
yet  to  be  found. 

Proceed  as  per  (8)  of  the  general  rule.  Take  the  sum  of 
the  first  and  second  terms  of  the  root,  or  7  for  the  value  of  a' 
or  m^  and  substitute  this  in  the  second  term  of  the  equation, 
(6;;2  +  4),  as  a  second  trial  divisor.  (6;;e  +  4)=6x  7  +  4=^46, 
the  second  trial  divisor. 

Divide  the  second  remainder  by  this  new  divisor.  This 
gives  1  or  ^  as  the  third  term  of  the  root.  Complete  the 
divisor  as  before  and  multiply  by  the  third  term  c.  This 
gives  [(6;;^  +  4)  +  3^]x^=[(6x7  +  4)  +  3]  Xl==49.  Sub- 
tracting leaves  no  remainder;  hence  the-root  is  3  +  4  +  1  or  8. 
Therefore  %=■%  or  x — 8^0. 

Divide  the  given  equation  Sx^'  +  ix — 224  by  this  quan- 
tity ;  this  gives  3;t:+28  or  3;r=— 28,  therefore  x=— 9/3  the 
other  root. 

Therefore  x=  |  _q^  and  the  factors  are  (x—S)  (3^-  + 28). 

In  this  example  we  have  three  terms  but  have  complied 

with  our  rule  for  the  extraction  of   a  root  and   taken  for  a 

second  trial  divisor  n  times  the  (n — 1)'^^  power  of  the  sum  of 


34  Solution  of  Equations, 

the  first  and  second  terms  of  the  root  viz,,  {a-^b)  or  (3  +  4) 
and  treated  this  as  one  term  and  proceeded  as  before. 

Example. 

Solve  the  equation  3^^— 2 jr=  43 2. 1468. 

In  this  equation  we  have  a  minus  sign,  coefficients  and 
decimals  to  deal  with.  Hereafter,  in  all  examples  of  more 
than  two  terms  in  the  result  or  root,  we  shall  consider  the 
second  and  successive  trial  divisors  as  n  times  the  {n — 1) 
power  of  the  sum  of  the  terms  already  found,  to  simplify  the 
solution  and  to  avoid  the  long  demonstrations  as  given  in  the 
example  on  page  (19). 

The  student  should  test  this  equation  for  positive,  nega- 
tive and  imaginary  roots  as  in  the  former  equations. 

The  student  can  readily  see  that  the  first  term  of  the 
required  root  may  be  given  any  value  whatever  ;  only  the 
greatest  value  that  can  be  readily  seen  to  satisfy  the  equa- 
tion is  preferable,  thus  reducing  the  number  of  terms  and 
the  labor  required  to  obtain  the  desired  result. 

It  is  probably  prudent  to  take  some  number  that  is  easily 
multiplied  and  divided  as  10  or  some  factor  of  10,  when  not 
too  great,  as  the  equivalent  of  a  or  the  first  term  of  the  root. 


abed 
4'32. 14^68  I  10+2  +  .3  +  .04 
2  80 


1  52.14  68=:lst  remainder. 
1  28 


24.14  68=2d  remainder. 
21.27 


2.87  68=3d  remainder. 
2.87  68 


Solution  of  Equations.  35 

Let  a=10,  then  3^^—2^=300—20=280 


1st  trial  divisor=6a^— 2=60— 2=58 
1st  complete  divisor=:=lst  T.  D.+2>b=^ 

58  +  6=64 
1st  complete  divisor  x  b  )  _^  .  ^  ^     ioq 
the  second  term        \  —^^ X  ^-li« 

Let  (a  +  d)=m 
2d  trial  divisor=6»^— 2=72— 2=70 
2d  complete  divisor=2d  T.  Z).+ 3^=70 +  .9=70. 9 

""      the  third  t'Irm     ^  '  j  =70.9 X  .3=21.27 

Let  {a  +  b-\-c)=n 
3d  trial  divisor=6;^— 2=73.8— 2=71.8 
3d  complete  divisor=3d  T,  Z>. +3^=71.8  +  . 12=71. 92 

^^  th^fourth  term  ^^  ^  =71.92  X  .04=2.8768 


10  +  2  +  . 3+04=12. 34 
;i:=12.34 
.^ir— 12.34=0 


^x^—        2.;*;— 432.1468  I    .^r- 12.34 


3x2_37Q2^  3:r+35.02 

35.02.r— 432.1468 
35.02.;t:— 432.1468 


3;c+35.02=0.  3:x:=— 35.02.  :v=  —  ^^=—11.67333  + 

o 

The  solution  of  the  above  equation  containing  decimals 
is  similar  to  the  former  solutions,  care  being  taken  to  comply 
with  the  negative  signs,  where  indicated,  also  to  observe 
decimal  quantities,  remembering  that  each  successive  decimal 


36  Solution  of  Equatio7is. 

term  is  given  its  proper  decimal  place  as  shown  in  the  example 
given.  No  further  detailed  explanation  is  necessary  if  what 
has  been  done  before  is  understood. 

In  this  example  x=^  \ ^^  *a^Q4-  ^i^d  the  factors  are 

(^—12.34),  (x+ll. 67333  +  ). 


CUBIC   EQUATIONS. 

A  cubic  equation  is  one  of  the  third  degree  as  ax^-\-bx'^-\- 
cx^=k. 

Solve  the  equation  x^ -\- x^  -\- x  —^\.%^=^ . 

In  this  equation  a  third,  a  second  and  a  first  power  of  x 
is  involved  and  the  sum  of  the  three  terms  is  equal  to  819. 
Hence  to  find  the  value  of  x  which  will  satisfy  the  equation, 
the  third,  the  second,  and  the  first  roots  of  819,  divided  by 
their  respective  coefficients  must  simultaneously  be  extracted. 

\.^\.x^{a-^b).  Therefore  ;»;3  +  ;t:^  +  jt:— 819  =  (^  +  ^)3  + 
{a-^by-^{a-^b) — 819=0.      Expanding  this  equals 

1  X  z 

1x2x3 
+  {^a"  +  lab  -\-b'')  =  second  term=i5  (^^-^  +  (?^— 1)  ^''"^^ 

-}-(^  +  ^)=third  term=C(^«-2+(;^— 2)^'^-3<^). 

Arranged  and  separated  according  to  (4),  the  equation 
equals  {a?-\-  a^  -\- a)  =  first  term  (Aa""  +  Ba""-^  +  C^«-2) 
+  (3a^  +  2«  +  l)/^=second  term  =\Ana''-^-\-B  iji—V)  ^^"^ 
+  C(7/-2)a«-3]^ 


Solutio7i  of  Equatu 


^1x2 


1x2x3 


1  X  z 


The  equation  now  assumes  this  form.  The  coefficients^, 
B,  Cin  this  equation  equal  1.  [a?-\-a^-^a)-^\{Za''-\-^a-\-V) 
+  3«^  +  ^^  +  ^]^^819. 

a  b 
819  I  5  +  4 
155 

664=lst  remainder. 
664 

Let  a^h,  then  a3  +  a2  +  ^:=53  +  5^  +  5=165 
1st  trial  divisor^3a^+2a+T-=75  +  10+l=-86 
1st  complete  divisor=rlst  T.  D.+Sad  +  d^  +  d= 
86  +  60  +  16  +  4=166 

'''    the  second  l.rm  ^J~l 66^^4^664. 

5  +  4-=9 

.T=9 

X— 9=0 

x^  +  x^  +  x—Sld  I     X—  9 


x^—9x'                     x^  +  lOx  +  dl 

lOx'+x 

91x— 819 
91;f— 819 

x'  +  \Ox=—U 

—91 

—25 

a            '    b 

(—6)  ±y—m 

—66 
66 

38  Solution  of  Equations. 

Let  a=— 6,  then  «2  + 10^=25— 50=— 25 
1st  trial  divisor  2a  +  l 0^—10+ 10=0 
1st  complete  divisor:rz=lst  T.  D.-\-b 

(0  +  ^)=/^ 
1st  complete  divisor   y.b\       7      /      /,         r>/> 
the  second  term  __  f  ^^  ^  ^-^^=-66 

Therefore  b=  +  ]/' — 66 

Explanation. 

Let  a  equal  some  number  whose  value  substituted  in  the 
above  equation  is  less  than  819. 

Let  a=5  as  a  trial  value  (5),  therefore  o? -\- a^ -\- a^=^b'^ -^-h"^ 
+  5=155.  Subtract  this  quantity  from  the  value  of  the 
given  equation  or  819;  this  gives  664  as  a  first  remainder. 

664  evidently  equals  [(3^=  +  2a  +  l)  +  3«/5  +  <^^  +  /^]^. 

The  trial  divisor  for  the  cube  root  is  3^=*,  the  trial  divisor 
for  the  square  root  is  2a,  and  the  trial  divisor  for  the  first 
root  is  1;  (all  these  comply  for  the  extraction  of  the  n^^  root, 
viz:  divide  the  first  term  of  the  remainder  by  n  times  the 
{n — Vf^^  power  of  the  first  term  of  the  root,  as  a  trial  divisor). 
3<^^  is  n  (n — ly^  power  of  a  where  ^/=:3  or  the  extraction 
of  the  cube  root;  2a  is  the  n  (n — ly^  power  of  a  where 
71=2  or  in  the  extraction  of  the  square  root;  1  is  the 
n  (n — ly^  power  of  a  where  ;2=1  or  the  extraction  of  the 
first  root,  which  is  simply  the  coefficient  of  a. 

3a^  +  2^  +  l=3x5^  +  2x5  +  l=86asthevalueof  the  trial 
divisor.  Divide  the  first  remainder  664  by  this  quantity. 
Let  this  equal  4  or  <^  the  second  term  of  the  root. 

Complete  the  divisor  by  adding  to  the  trial  divisor  the  nec- 
essary quantities  for  the  completion  of  the  divisors  in  the  ex- 
traction of  the  cube  and  square  roots;  that  is,  add  Sad  +  d^=S 
X  5x4  +  4x4  for  the  completion  ot  the  cube  root  divisor, 
and  ^  or  4  for  the  completion  of  the  square  root  divisor,  thus 
making  the   complete    divisor,    l{Sa''  +  2a  +  i)  +  Sad  +  d''  + 


Solution  of  Equations.  39 

^]==86  +  80=166,  and  multiply  this  by  4;  this  gives  664. 
Subtract  this  from  the  first  remainder,  which  leaves  no 
remainder;  hence  the  root  is  5  +  4=9  or  x=9, 

X — 9=0  is  therefore  a  factor  of  x^  +  x^'+x — 819=0. 

Dividing  x^  +  x^  +  x—S19  by  (x—d)  gives  jt:=  +  10jtr+91 
as  the  other  root  of  the  equation  x^  +  x'^+x — 819=0=(^ — 9) 
(x^  +  10ji:  +  91). 

We  must  now  comply  with  the  method  of  solving  the 
equation  of  the  second  degree  to  find  the  other  factors. 

In  the  equation  x^+  10;t:+91=0,  we  have  two  permanences; 
hence  the  two  roots  are  negative.  A"" — 2B  is  negative,  hence 
both  roots  cannot  be  real. 

Let  x=(a  +  d)  as  before  and  solve  for  the  remaining  roots. 

x'  +  10x-\-n=(a  +  dy  +  10(a  +  d)  +  91^0^a'^  +  2ad  +  d^  + 
10^  +  10^  +  91=0=  (a^  + 10a)  +[(2^  +  10)  +  ^]  ^=—91. 

Assume  a^= — 5,  since  the  root  is  negative,  therefore  a^  + 
10^=25—50=  —25.  Subtract  this  from  —91,  leaving  —66 
the  first  remainder.  2^  +  10  the  first  trial  divisor  equals 
—10  +  10=0. 

—66  evidently  equals  [(2a  +  10)  +  d']  b\  but  2a +  10 
equals  0;  hence  (0+^)/^  or  ^^^  — 66  or  /5=  +  ]/— 66  an 
imaginary  quantity  as  shown. 

(      ^        — 

Therefore  x=J  — 5  +  |/— 66 

I  _5_^/Z:66 

By  testing  before  solving  the  equations  for  positive  and 
negative  roots  and  imiginary  quantities,  according  to  corillary 
(6),  the  student  will  be  somewhat  guided  in  the  selection 
of  terms  for  a  and  b  and  avoid  what  may  lead  to  tedious  work 
and  confusion. 


40  Solution  of  Equations, 

Example. 

Solve  the  cubic  equation  x'^^-^x^ — 37^—210=0. 

The  roots  of  this  equation  are  three  in  number,  two  nega- 
tive and  one  positive  ;  and  all  real.  (The  student  is  requested 
to  test  each  equation  for  real  and  imaginary  roots,  also  fo^ 
positive  and  negative  roots  by  the  theory  given  on  page  24). 

Let  x^=^(a-\-b)  as  before. 

(1).  Therefore  x'^-\-  6  x^ — ^7  x  —  210  =  0,  substituting 
(a  +  d)  equals 

(2).  (a  +  dy  +  6  (a  +  dy—S7  («  +  ^)— -210=0.  Expand, 
ing  equals  (a^  +  Sa^d+Sad^  +  d^)  +  6  {a^  +  2ab  +  b^)—^l  {a  + 
^)— 210=0. 

Separating  and  arranging  according  to  (2),  (3),  and  (4)  of 
the  rule. 

(3).  (a3  +  6^2__37^^)  +  |-(3^2^  12a— 37)  +  3a^+^^  + 
6^]  ^=210. 

:r3  4- 6jt:^— 37^=210 

a  b 
210  I  4  +  2 
J_2  "~" 

198=lst  remainder. 
198 

Let  a=^,  then  a^  +  ^a'—Zla^^^-^m—l^&^ll 


1st  trial  divisor=3a^  +  6x  2a— 37= 

48  +  48-37=59 
1st  complete  divisor=Ist  Z!  D.-\-Zab-\-b^-{-^b^ 
59+24  +  4+12=99 

the  second  term         |  '""^^  ^  2=198 


Solution  of  Equations.  41 

4  +  2==6 
x=^ 
:r— 6=0 
x3  +  6jt:^— 37;t:— 2101  JIT— 6 


x^—^x^      _^___    x^  +  12x+S5 
12x^—Zlx 
12x^—12x 


35x— 210 
36.r— 210 


X  +12;tr=— 35 

a     b 

— 35| 

—4—1 

—32 

—  3 

—  3 

Let  a— 

—4,  then  a^+12a= 

=  16—48=32 

1st  trial  divisor=2a+ 12=— 8  +  12=4 
1st  complete  divisor==lst  T.  Z^. +  /^=4  +  (— 1)= 
1st         *'  '*      X. 

the  second  term 


'  i  Z..3  X  (— 1)=— 3 


(_4)  +  (-l)=-5 

x=—h 

Jir  +  5=0 

x^+12a:+35 

x  +  5 

x^  +  5;«r 

x->rl 

7;t:+35 

x+7=0 

7a- +  35 

;.:-=— 7 

Hence  the  factors  equal  {x — 6)  (ji;  +  5)  (jir  +  7)- 
Explanation. 

For  a,  substitute  some  numerical  trial  value,  whose  value 
substituted  in  the  first  term,  {a}-\-^a^ — 37«),  is  less  than  210. 
Assume  a  equals  4.     (If  4  should  satisfy  the  equation,  we 


42  Solution  of  Equations. 

would  have  only  one  term  in  the  factor,  and  this  substituted  in 
the  equation  would  satisfy  the  equation  at  once;  hence  there 
would  be  no  necessity  for  a  second  term  b,  and  a}-\-^a'^ — 
37«  would  equal  210;  but  if  a  does  not  satisfy  the  equation 
we  have  still  a  second  or  perhaps  a  third  term,  and  so  on 
until  we  have  reached  a  desired  root,  if  it  be  a  commensur- 
able quantity),  {a} -\- ^a"  —  Zld)  therefore  equals  (43  + 
6x4^—37x4)^12. 

Subtract  this  from  the  absolute  term  210,  leaving  a  remain- 
der, 198.  198  evidently  equals  the  remainder  of  the  equa- 
tion [  (3a^  +  l2a— 37)  +  3a^— /^^  +  6^]^.  (3),  after  the  first 
term  a^  +  Qa"" — S7a  have  been  eliminated. 

Now  observe  that  (Sa''  +  12a — 37)  is  the  first  trial  divisor. 

Sa^  is  the  trial  divisor  for  the  cubic  portion, 
6x2a    "    *'     *'  ''       ''     ''    quadratic  portion, 

— -37x    a°  ''    ''     ''         ''       *'      ''    first  power  portion; 

hence  the  sum  is  the  complete  trial  divisor,  and  all  are  the 
ti  (n — ly^^  powers  of  the  highest  powers  of  a  according  to 
rule. 

3a»  +  6 x  2a— 37==  3x4^  +  6x2x  4—37=59,  the  numeri- 
cal trial  divisor.  Divide  the  first  remainder  by  69,  let  this 
equal  2=5.  (The  student  may  try  2  by  substituting  its  value 
in  the  complete  remainder  to  see  whether  the  quantity  be  too 
great,  just  as  in  the  extraction  of  any  root). 

Complete  the  divisor  by  adding  the  remainder  of  (3)  to 
the  trial  divisor  and  substitute  the  values  of  a  and  d  in  the 
equation  and  multiply  by  <^==(lst  T.  D.+  Sad+  d'^+Qd) 
^=:(59  +  3x  4x  2  +  2^  +  6x  2)x  2=99x  2=198;  this  leaves 
no  remainder. 

Hence  x==6  and  x — 6^0,  one  of  the  factors. 

Divide  ;t:3  +  6jt:^— 37jt;— 210  by  (:r— 6)  rule  (3),  this 
gives  ^^ +  12^+35.     (4). 


Solution  of  Equations.  43 

Solve  this  equation  (4)  to  determine  the  other  roots  or 
factors.     Solving  as  shown  gives  x=  — 5  a  negative  root  or 

Divide  x^-^Vlx-\-Zh  by  x-i-5,  this  gives  ^+7,  the  other 
factor  as  shown. 

(   ^ 

We  now  have  the  three  roots,  ;f=  -\  — 5  or  two  negative  and 

(-7 
one  positive  root. 

The  factors  are  {x — 6),(jt:+5),(-r  +  7),  therefore  (jr — 6) 
Cr+5)  (x  +  7)--;tr3  +  6jt:^— 37:tr— 210=-0. 

Example. 

Solve  the  equation  2^^  _g5o^.^3000. 

This  equation  has  no  second  power  of  x,  hence  it  is  not 
complete;  the  second  term  may  be  supplied  thus,  2;t:^  + 
Ojt^— 650ji-=:3000. 

The  solution  of  this  equation  is  solved  like  the  former 
equations,  only  there  being  no  term  of  the  second  degree  in 
this  equation  there  is  of  course  no  trial  nor  complete  divisor 
for  the  second  degree  extraction;  hence  simply  deal  with  the 
third  and  first  degrees  as  shown.  This  example  needs  no 
further  explanation. 


44 


Solution  of  Equatio7is, 


be 


,      « 

+   CD 

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vi 

5  ^ 

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lO      1 

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(N 

a  S 

a>  CO 

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^   <s 

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1      1 

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Solution  of  Equations. 


45 


4- 

o 


+ 


T  + 


^^  o 

ST 
5^ 


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h4  11 


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'd 

□ 

<:>  <M 

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1 

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00 

(M 

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46  Solution  of  Equations. 

If  the  second  term  of  function  x  does  not  appear,  the  sum 
of  the  roots  of  the  equation/  (x)=0  is  zero.  Thus  the  sum 
of  the  roots  of  the  above  equation  equals  Ot  viz.  — 20  +  5  + 
15  equals  0  from  (3)  page  (22). 

Example. 

Solve  the  equation  3;t:3—4;i-^+6;tr— 82068.904448=0. 

In  this  equation  we  have  negative  and  positive  signs  and 
decimals.  The  solved  example  will  show  all  that  is  neces- 
sary without  any  further  explanation. 

3;«:3—4;tr»  +  6;r  =  82068.904448 
I^eta:=(«  +  ^),  then  %x^—^x^  ^iSx  =  Z  (a  +  ^)3— 4  (^  +  <^)"  +  6  (^  +  <5)  = 
3  («5+3«»/^  +  3«32+33)_4(^2+2a3  +  ^'')  +  6  (^  +  3)  = 
3fl 3-_4^ 2  4-6^+  [(9a^— 8^  +  6)  +  ^ah-\-U "— 4<5] h 

82068.904448  |  20  +  5.8  +  4.76 
22520 


59548.904448=  1st  remainder. 
26492.776 

33056.128448  =  2d  remainder 
33056.128448 
lyet  a  =  20,  then  Za  3— 4a''  +  6^  =  24000—1600  + 120  =  22520 

1st  trial  divisor  =  9a ^ — 8a  +  6  = 

3600—160  +  6  =  3446 
1st  complete  divisor  =  1st  T.  D.  +9a^  +  33='--4^  = 

3446  + 1044  + 100.92—23.2  =  4567.72 

'^^tgco^g^^  '  \  =4567.72x5.8  =  26492.776 

Let(a  +  3)  =  w 
2d  trial  divisor  =  ^m'^—'^m  +  6  =  5990.76—206.4  +  6  =  5790.36 
2d  complete  divisor  =  2d  T,  D.  +  ^mc  +  Zc^ — Ac  = 

5790.36  + 1105.272  +  67.9728—19.04  =  6944.5648 

^^  Te^lwrd^ilrm'  ""  ^  }=  6944.5648  x  4.76  =33056.128448 

Therefore  ^  =  20  +  5.8  +  4.76  =  30.56. 

The  student  is  requested  to  test  and  solve  for  the  remain- 
ing roots.     There  are  three. 


Solution  of  Equations.  47 

Example.  ^^ 

Solve  the  equation  x"^  +  llji;3—42x='—680jt-— 1600^0.  1 

In  this  equation  there  are  three  permanences  of  signs  and 
one  variation,  hence  by  Descarte's  rule  of  signs  there  will  be     ^ 
one  positive  root  and  three  negative  roots.  '^     ^ 

The  roots  of  the  equation  are  all  real  because  A"^ — ^B  is     §  ^x 
positive.     121 — ( — 84)  equals  a  positive  quantity.  '^  ^  | 

Let  x=^{a-\-U)  as  in  former  equations.  Therefore  x"*  +  lljr^     I  ^ 
—42^1:^— 680-r  — 1600-=  (a  +  /5)*  + 1 1  («  +  Uf—  42(a + /^)»—  V 
680(a  +  ^)— 1600=0  -^  V 

(a*  +  4.aH + Wb^  X  ^ab^  +  b^)  + 1 1  (^3  +  Za^b  +  Zab^  +  ^3)^42 
(^^  +  lab^  ^0— 680(«+ ^)— 1600=0, 

Arrange  and  separate  the  terms  so  that  all  the  powers 
of  a  beginning  with  the  highest  power  are  consecutive:  so 
that  all  the  terms  used  in  the  trial  divisor  of  the  various 
powers,  beginning  with  the  highest  power,  are  consecutive  : 
so  that  the  remaining  terms  arranged  according  to  their 
respective  positions  in  the  general  equation  according  to  rule. 
(^*  +  ll«3+42a:^_680«)  (1.) 
+  [(4^3  +  33a-~84«— 680)  (2.) 
+  6a^^+4«^^  +  <^3  +  33a^+l  1^^—42^]^     (3.) 

We    will    solve   for   the  positive   root    first  :    since   there 
are  one  positive  and  three  negative  roots  :  let  x  =:4:^=a  the 
first   term  of  the  positive  root,  therefore  a'^  +  lla^ — 42a^ —     f 
680a=— 2432  as  shown.     Subtract —2432  from  1600:  this    J 
gives  4032  the  first  remainder  (2)  and  (3)  above.  -  < 

Take  the  algebraic  sum  of  the  various  trial  divisors  of  the     |  . 
various  powers"  of  a  as  the  complete  trial  divisor  shown  in  (2 j      ^^ 
above  :  each  term  of  which  is  equal  to  the  n(n — l)tk  power     1 
of    a,    multiplied    by   their    respective    coefficients :    viz.,    ^ 
4^3 + 33^^—84^—680=256  +  528—336—680=  —232    the  ^ 
first  trial  divisor. 


48  Solution  of  Equations. 

Divide  4032  by  the  trial  divisor;  this  equals  b  or  4,  the  sec- 
ond term  of  the  root.  Complete  the  divisor  by  adding  to  the 
trial  divisor,  this  value  for  ^  or  4  substituted  in  (3)  above, 
this  gives  1008;  multiply  this  by  4,  which  gives  4032. 

Subtract  this  from  the  first  remainder  leaving  0 :  therefore 
4  +  4=8  is  a  root  of  the  equation  or  ^=8.  The  factor  is 
therefore  x — 8  or  x — 8=r0. 

The  remaining  roots  are  found  as  shown  by  the  example 
and  are  of  course  similar  to  former  cubic  and  quadratic 
equations. 

The  student  should  test  each  equation  for  positive  and 
negative  roots,  and  for  real  and  imaginary  roots,  to  become 
familiar  with  the  use  of  the  given  theory. 


Solution  of  Equations. 


49 


o 


a 


CO    CO 

o  o 


+ 


o 

(^ 

C<l 

s 

4- 

1-1 

^ 

II 

o 

1—1 

H 

rH 

+ 

1 

^ 

1— ( 

H 

°f 

+ 

I 

+ 

crr 

i 

1^ 

+ 

tH 

<i 

+ 

(M 

Q 

<M 

2 

II 

CO 

+ 

CO 

CO 

1 

H 

^ 

II 

T 

8 

o 

1 

QO 

1 

-C 

r-K 

II 

CO 

OO 

II 

1 

II 

•^ 

00 

' 

1 

g 

II 

S 

X 

T 

1 

Q 

^ 

+ 

rH 
1 

§ 

1 

CO 

I 

CO 

ci 

^ 

1 

CO 

o 

1 

1 

1 

CO 

1 

t>. 

11 

•^ 

m 

Q 

J3 

?3 

CO 

1 

(-; 

1 

4- 

J 

"H 

1— t 

(N 

£2 

00 

CO 
rH 
II 

tH 

00 

-o 

"^ 

CO 

(M 

1— t 

(M 

00 

+ 

1 

+ 

>o 

+ 

IC 

X 

»— 1 

<2 

II 

1 

II 

CO 

8 

+ 

1 

H 

^ 

1 

u 

'X> 

• 

ft 

II 

1 
1 

o 

''5 

P. 

a 

o 
y 

4-» 

+ 
CO 

i 

O 
+ 

5S 

3 

■M 

+ 

85 

b^ 

r-H 

CO 

T— 1 

QQ 

1— t 

H 

^ 

o 

00 

1 

t 

l^ 

1 

1 

o 

o 

T— 1 

1 

1—1 
1 

tH 

1 

1 

"H 

'5^ 

a> 

a> 

T— < 

rH 

o  o 

CO    CO 


50 


Solution  of  Equations. 


T^^ 


li    + 


I       I 


1  H   H 


2    2 


+ 

CO 

s 
^ 

+ 

II 

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1 

"^ 

CO 

1 

1 

II 

5S 

CO 

+ 

1 

CO 

1 

?s 

II 

T-\ 

1 

o 

O 
rH 
rH 

Q* 

+ 

X 

+ 

+ 

^ 

+ 

II 

^ 

.1 

-4-> 

O 

T— 1 

Oi 

CO 

(fl 

<i 

rH 

+ 

r-K 

1 

+ 

II 

X 

?s 

o 

CO 

u  a 

Q 

CO 

O    U 

II 
.2 

CO     C> 

*> 

1^ 

f> 

<P 

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4-> 

■4-»  a 

f 

*^ 

(U 

1 

, 

Ph 

cfl 

a 

« 

s 

8 

o  ti 

+ 

+ 


+    + 


SSI 

(N    OQ  1 

% 

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+  •+ 

rH 

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o  p 

rH 

CM    C^ 

+ 

+ 

'^ 

H 

^ 

OS 

I  I 


-"^ 

lO 

-f 

+ 

H 

^ 

o 

(M 

+ 

H 

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a> 

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(M 

+ 

+ 

II 

N 

CO 

rH 
1 

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o 

7 

1 

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1 

1 

11 

1 

II 

H 

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+ 

1 

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1 

+ 

II 

c:5 
+ 

q' 

X 

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CO 

1 

II 

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to 

rH 

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II 
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a 

+ 

a; 

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o 

a 

5 

11 

CO 

CO 

1 

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ii 

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a 
o 

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0) 

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s 

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C<l 

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h4      ,-(         — 1    '  rH 


Solution  of  Equations.  51 

Example. 

Solve  the  equation  2jt:5  +  3jt*— 220Jl:3  +  84;f^  +  4880Jt;— 
9600-0. 

In  the  above  example  we  have  an  equation  of  the  fifth 
degree;  hence  five  roots  or  factors  are  available.  By  Des- 
carte's  rule  there  are  three  positive  and  two  negative  roots. 
A^ — IB  is  a  positive  quantity  hence  all  the  roots  are  real. 

Arrange  and  separate  the  terms  so  that  all  the  powers  of  a 
beginning  with  the  highest  power  are  consecutive;  so  that  all 
the  terms  used  in  the  trial  divisors  of  the  various  powers 
beginning  with  the  highest  power  are  consecutive;  so  that  all 
the  remaining  terms  are  arranged  according  with  their  re- 
spective positions  in  the  general  equation. 

Let  x^r^{a-\-b).  Substitute  this  value  in  the  given  equation. 

(1).  2(^  +  /^)5  +  3(a  +  ^)4— 220(^  +  ^)3  +  84(^  +  ^)^—4880 
(a  +  ^)-=9600. 

(2).    2(a5  +  5«^^  +  10«3^^  +  10^^^3  +  5^^4  +  ^5) +3(^4  +  4^3 

^  +  6^^^^  +  4^<^3_|_^4^_220(a3  +  3a^^+3a^^  +  ^2)  +  84(a^  + 
lab  +  b-^  -4880(«  +  ^)=9600. 

Arranged  and  separated  according  to  rule,  this  equation 
assumes  this  form,  viz.: 

(3).    (2^5  +  3^*— 220^3 +  84^^— 4880a) -r        first  term. 
+  [(10a'^  +  12a3_6eoa2^1g8^__48gO)^     -        second  term 
+  (20^3^  +  20a^^'  +  10a^3  _|.  2^^  +  \U^b  +  Vlab^  +  3^3«-660a^ 
— 2203^  +  84<^)]/^=:third  term 
==9600=absolute  term. 

The  first  term  of  (3)  consists,  evidently  of  the  highest 
powers  of  a  in  each  consecutive  term  of  the  expansion  (2). 
The  second  term  of  (3)  is  the  complete  trial  divisor,  or  the 
n{n — \.y^  powers  of  the  consecutive  terms  of  (1).  The 
third  term  is  the  remainder  of  the  above  expanded  equation, 
or  the  quantity  added  to  form  the  complete  divisor  and  the 
sum  multiplied  by  the  second  term  of  the  root  b,  as  shown. 


62  Solution  of  Equations , 

The  solution  is  the  same  as  in  former  problems. 

In  this  example  we  have  solved  for  one  of  the  positive 
roots  first  which  equals  4  as  shown;  and  the  factor  is  x — 4- 

Divide  the  first  equation  by  this  factor  which  gives  2x''-\- 
Ilx3_i76^2_620^+ 2400=0. 

Solve  as  before  for  the  next  root.  This  gives  the  negative 
root  ■ — 6  and  the  second  factor  equals  (jr  +  G). 

Divide  the  equation  2;t:*+ll:v3— 176x^—620^ +2400=0 
by  the  factor  (.r  +  G),  this  gives  the  cubic  equation  or  por- 
tion of  our  required  solution.  Proceed  as  before  until  all 
the  desired  roots  and  factors  are  obtained. 

We  find  the  factors  to  be  (jtr— 4)(x  +  G)(;r— 8)(2j»:— 5) 
(x+10). 

"    X=4: 

x=—6  .     ^ 

The  roots  are  -!   x=S 

x=— 10 
2x=6 

For  equations  of  higher  degree  proceed  as  in  the  example 
given,  care  being  taken  for  the  systematic  arrangement  of 
terms,  trial  divisors,  etc. 


Solution  of  Equations. 


53 


+ 


o 

o  vo 

II 

822 

o 

a.  o 

o 

CD 

Oi 

I 

I 

00 


ou 

+ 

CI 

I 

CO 


C<l 


S 


i; 

I  I 


II 

+  ^ 

Q  + 

I  + 

CO  05 

<u  + 

;d  CD 

-M  00 


J3 


pH  1— t  lO    tH 


54 


Solution  of  Equations. 


1  ^  H 


I 

I 


H 
o 

I 




II 

o 

CO 

(M 

C<J 

<M 

t^ 

t- 

lO 

CO 

00 

1 

1 

T 

7 

CO 


II 


II 

CO    ^ 

Is 

^    I 


a 


a; 


II 
o  ^ 

is 

CO      I 
CO 

CO     + 

II   c^ 
o  ^ 


CO 

CO 
Oi 

n 


kj  -ci  T^l 

CO  CO 

^    n"  + 

"^    J  2 


II      ,-H 

o    + 

•>   « 

(U 

•4-* 

(U 

»— t 

a. 

a 

o 


05 
II 


<b 


c^    o  S 

CO     <u    o 


+ 


CM 

+ 

o 


N 

7 

"^ 

'^ 

y-K 

(M 

r-\ 

rH 

+ 

-I- 

'S- 

^ 

^ 

H 

(M 

(N 

H 

^ 

o 

o 

(M 

(M 

1 

O 
1 

^ 

N 

^1 

1 

1^ 

H 

i^ 

H 

CO 

CO 

o 

o 

h>. 

t^ 

I-- 

T 

1 

1 

rH 
1 

! 
m 

1 

1 

8| 

S    C<l 

+  + 

H   H 

§  S 

SoluHofi  of  Equations. 


65 


+ 


O     00  I  C<J    ^  i 
O     (M     t^    t^ 

I         I         I        I    I 


il  iC  II 

^  C^  lO  o 

+  11  II  I 

(M  ^  H  H 


I 


CO 

1 

1 

11 

1 

II 

<i 

1! 

o 

1 

(M 

o 

t^ 

i 

t^ 

^ 

»— t 

•c> 

1 

CO 

1 

<M    ^ 

1 

T 

II 

1 

-^ 

.S 

X 

cr 

<M 

CO      1 

+  11 

Tt" 

rH 

II 

Tfl 

1! 
O 

O 

rH 
1 

Si  ^, 

1—1 
1 

II 

tH 

« 

+->   + 

, ' « 

1 

7 

^=o 

^li 

1 

1 

1 

II  + 

X 

M 

ss 

s  ^ 

« 

II 

•^    1 

•Ih    -^ 

<M 

J5 

>    1 

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C 

> 

*'d 

^d   0 

Si 

ii  § 

-«-» 

'^ 

<u 

C^^ 

P. 

P-i  1) 

II 

3 

a 

a  :2 

Q 

o 
o 

o  *" 
o 

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^ 

•4-> 

4-> 

CO 

rH 

CO 

rH 

en 

T-l 

o 
o 

+ 

o 


H  V5 
O  O 
!>.    i-H 


+     + 

o 


H     ^ 


56 


Solution  of  Equations. 


II 


(M 

+ 
:    CD 


O    00 
00    TT 


a 


52    (M 

CO    CO 


O 
II 

I  I 


(M     00 

+    I 


u 

o  o 


i 

1 

Cm 

CD 

+ 

H, 

1 

1 

OJ 

u 

aJ 

to 

o 

Vi 

rH 

3 

+ 

a 

^ 

0) 

00 

rd 

1 

-♦-• 

1 

TJ 

^ 

a 

C(J 

rf    CO    U5 

00   o 

1     ^ 

T 

^ 
V 


Solution  of  Equations,     \  ^J  ^^^Tf 

Example. 

Find  the  roots  of  x  in  the  equation  6x^-{'12mx^ — dnx" —  t 

lSm7ix  +  4kx^  +  Smkx+6nkx — 12mnk=0.  ^f] 

The  above  equation  is  an  equation  of  the  third  degree,  or  ^  | 
a  cubic  equation,  where  6,  12m,  dn,  ISmn,  etc.,  are  "l^  ^ 
coefficients  of  jtr  corresponding  to  the  coefficients  A,  B,  C,  .5  '^ 
Dy  etc.,  of  the  general  equation  (1).  The  exponent  is  three  ^;| 
and  the  absolute  term  is  \2nink\  each  respectfully  corres-  ^'^ 
ponding  to  the  exponent  n  and  the  absolute  term  K  of  the  jj  ^ 
general  equation  (1).  ^^ 

The  roots  of  an  algebraic  equation   are   functions  of  its     |  >^ 
coefficients;  hence   the  roots  of  this  equation  are  evidently 
w,  n  and  k  combined  with  the  numerical  terms  of  the  given 
equation.  ^  A     % 

In  the  above  equation  let  x=^{a-\-b),  where  a  and  b  are  ^  | 
terms  of  the  desired  root.  Substitute  this  value  in  the  given  5  ? 
equation  and  expand  according  to  (1)  of  the  rule.  >  "^ 

(V)  6x^  +  12mx^—9nx'—lSm?ix  +  4kx^  +  Smkx+enkx^  -i 
— 12mnk=0.  c^  4 

(2)  6x^  +  (12m—dn  +  ik)x^  +  (Smk—lSmn—6nkyv-~'^  | 
12mnk=0.  ^  • 

(3)  6(a  +  d)^-{-(il2m-~dn  +  4k)  {a  +  by  +(i^mk—\^mn  ,^ 
—^nk)  (a  +  b)  =12mnk. 

The  expanded  form  equals 
6  (a^  +  Za^b+^ab^  +  b^)  =-lst  term  \ 

+  {12m—9?i  +  4k)  (a^  +  2ab-\-b')=^2d  term  l=12mnk=(4) 
+  {Smk—lSmn—6nk)  (a-\-b)     r^3d  term  J 

Separating  and  arranging   (4)  according   to  the  rule   (4)  O 
equals 
({^a^  +  a%12m—9?i  +  4k)  +  a  {^^mk—\%mn'\  -^ 

— ^nk)  =lst  term 

+  [(18a^  +  2a{\2ni—%i  +  4k)  +  (8  w/^— 

X'^mn — ^7ik)^  \  =2d  term 

-^\%ab^U^^b[i2m—9n-V4k)~\b=Zd\.^xm^ 


Y 

\2mnk=(h)  ^     3^ 


58  Solution  of  Equations. 

^a^  +  a^{\2m  —  2n^4:k)  +  a{^mk—lSm7i—Q7ik)  are  the 
terms  separated  and  arranged  so  that  all  the  highest  powers 
of  a  in  each  consecutive  term  of  the  expanded  form  (4)  are 
consecutive.  These  terms  correspond  to  the  A a^-{-Ba^~'^+ 
Ca^~'^f  etc.,  of  the  general  equation. 

lSa''  +  2a(12m—dn  +  4k)  +  (Smk—lSmn—6nk)  are  each 
respectfully  the  n  times  the  (n — 1)  power  of  the  descending 
powers  of  «  and  correspond  to  the  Ana^~^-{-B(n — l)a^~^+ 
C{n — 2)^^~^  of  the  general  equation;  that  is  Za'^  is  the 
{n — X)*^  power  of  <^,  this  times  its  coefficient  6  equals  18a=^; 
2a  is  the  n  {n — V)^^  power  of  a,  this  times  its  coefficient 
equals  2a  {Vim — dn  +  4k);  a'  or  1  is  the  n{n — l)^'^  power  of 
Uy  this  times  its  coefficient  equals  (Smk — ISmn — 6?ik);  the 
sum  of  these  is  the  complete  trial  divisor  for  the  finding  of 
the  remaining  terms  of  the  desired  root  and  each  is  respect- 
fully the  trial  divisor  for  the  simultaneous  extraction  of  the 
cubic,  quadratic  and  the  first  power  root  as  indicated  by  the 
equation. 

The  remaining  terms,  lSad  +  6d''  +  d(12m — 9n-\-4:k),  evi- 
dently must  be  added  to  the  trial  divisor  to  form  the  complete 
divisor  after  the  second  term  d  of  the  root  has  been  found. 
This  complete  divisor  multiplied  by  the  term  d  completes  the 
extraction  of  the  required  root  if  there  be  no  remainder.  If 
there  be  a  remainder  let  the  sum  of  (a-i-d)=a^  s-nd  proceed 
as  before. 

By  referring  to  the  example  given  and  solved  we  have  let 
a= — m  and  substituted  it  in  the  first  term  of  (5)  which  equals 
the  quantity  shown.  Subtracting  this  from  the  absolute  term 
12mnk  equals  the  first  remainder. 

The  trial  divisor  is  equal  to  the  second  term  of  equation 
(5j  with  the  value  of  — m  substituted  for  «,  which  gives  the 
quantity  shown  and  marked  as  first  trial  divisor. 

Divide  the  first  term  of  the  first  remainder  or  the  remain- 
ing portion  of  the  absolute  term   by  the  term  of  the  trial 


Solution  of  Equations,  59 

divisor  that  contain  the  same  factors;  that  is,  follow  the  rule 
of  algebraic  division.  The  quotient  is, the  second  term  of 
the  root  or  b,  (^mnk  divided  by  —  6/^/^= — m,  the  second 
term  of  the  root.) 

Complete  the  divisor  by  adding  to  the  trial  divisor  the  third 
term  of  equation  (5),  where  b=^ — m^  and  multiply  the  com- 
plete divisor  by  b  or  its  value  — m  as  shown.  This  quantity 
subtracted  from  the  first  remainder  leaves  0;  hence  the  equa- 
tion is  solved  for  one  of  the  roots  which  equals — ^m  and  the 
factor  is  (;i;  +  2;«). 

Divide  the  given  equation   by  this  factor;  this  gives  6jr^ — 

dnx-\-4kx — 6?ik=0.     In  this   equation    proceed   as   before 

3 

and  extract  the  root  n  as  shown.     The  factor  is  (x — -;2), 

Ji 

or  (2x — Sn). 

Divide  the  equation  6x'' — 9nx  +  4kx — 6?ik  by  (2x — 3;^); 
this  gives  Zx-\-1k  the  third  factor. 

Therefore   6^;^+  Vlmx"^  —  ^nx"^ -\- 4kx'^ —  \.%mnx -k-^mkx 
+  6nkx — Vlmnk  equals  (;i;  +  2;;e)  (2x — Zn){^x-\-1k), 
j  — 2m 
\      3 
The  roots  are    (       2 


60 


Solution  of  Equations, 


n3 
S 


II 

<^ 

<^ 

•^ 

« 

s 

^ 

s 

TJ^ 

-^ 

-^ 

1 

+ 

+ 

8 

8 

8 

§ 

§ 

1 

05 

05 

Oi 

+ 

J 

J 

^ 

1 

1 

CO 

<X) 

CO 

+ 

1 

1 

<^ 

►^ 

<ii 

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Solution  of  Equations. 


61 


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XI 


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